THE  MAGNETIC  CIRCUIT 


WORKS  BY   THE   SAME  AUTHOR 


Published  by  McGRAW=HILL  BOOK  COMPANY 

The  Electric  Circuit 

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The  Magnetic  Circuit 

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Cloth net,    $3.50 

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Engineering  Applications  of  Higher  Mathe 
matics 

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THE 

MAGNETIC    CIRCUIT 


BY 

V.   KARAPETOFF 
M 


MCGRAW-HILL  BOOK   COMPAiSTY 

239  WEST  39TH  STEEET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.G. 

1911 


TK  /.T3 


Engineering 
Library 


Copyriglit,  1911 

BY 

MCGRAW-HILL  BOOK  COMPANY 


PREFACE 


THIS  book,  together  with  the  companion  book  entitled  "  The 
Electric  Circuit,"  is  intended  to  give,  a  student  in  electrical 
engineering  the  theoretical  elements  necessary  for  the  correct 
understanding  of  the  performance  of  dynamo-electric  machinery, 
transformers,  transmission  lines,  etc.  The  book  also  contains 
the  essential  numerical  relations  used  in  the  predetermination  of 
the  performance  and  in  the  design  of  electrical  machinery  and 
apparatus.  The  whole  treatment  is  based  upon  a  very  few  funda- 
mental facts  and  assumptions.  The  student  must  be  taught  to 
treat  every  electric  machine  as  a  particular  combination  of  electric 
and  magnetic  circuits,  and  to  base  its  performance  upon  the 
fundamental  electromagnetic  relations  rather  than  upon  a  sepa- 
rate "  theory  "  established  for  each  kind  of  machinery,  as  is  some- 
times done. 

The  book  is  not  intended  for  a  beginner,  but  for  a  student 
who  has  had  an  elementary  descriptive  course  in  electrical  engi- 
neering and  some  simple  laboratory  experiments.  The  treat- 
ment is  somewhat  different  from  that  given  in  most  other  books 
dealing  with  magnetic  phenomena.  It  is  based  directly  upon 
the  circuital  relation,  or  interlinkage,  between  an  electric  current 
and  the  magnetic  flux  produced  by  it.  This  relation,  and 
the  law  of  induced  electromotive  force,  are  taken  to  be  the 
fundamental  phenomena  of  electro-magnetism.  No  use  what- 
ever is  made  of  the  usual  artificial  concepts  of  unit  pole,  magnetic 
charge,  magnetic  shell,  etc.  These  concepts  of  mathematical 
physics,  together  with  the  law  of  inverse  squares,  embody  the 
theory  of  action  at  a  distance,  and  are  both  superfluous  and 
misleading  from  the  modern  point  of  view  of  .a  continuous  action 
in  the  medium  itself. 

The  ampere-ohm  system  of  units  is  used  throughout,  in 
accordance  with  Professor  Giorgi's  ideas,  as  is  explained  in  the 


254299 


vi  PREFACE 

appendices.  Those  familiar  with  Oliver  Heaviside's  writings 
will  notice  his  influence  upon  the  author,  in  particular  with 
regard  to  a  uniform  and  rational  nomenclature.  The  author  trusts 
that  his  colleagues  will  judge  his  treatment  and  nomenclature 
upon  their  own  merits,  and  not  condemn  them  simply  because 
they  are  different  from  the  customary  treatment. 

In  the  first  four  chapters  the  student  is  introduced  into  the 
fundamental  electromagnetic  relations,  and  is  made  familiar  with 
them  by  means  of  numerous  illustrations  taken  from  engineering 
practice.  Chapters  V  to  IX  treat  of  the  flux  and  magneto- 
motive force  relations  in  electrical  machinery,  first  at  no  load,  and 
then  under  load  when  there  is  an  armature  reaction.  The  remain- 
ing four  chapters  are  devoted  to  the  phenomena  of  stored  magnetic 
energy,  namely  inductance  and  tractive  effort.  The  subject  is 
treated  entirely  from  the  point  of  view  of  an  electrical  engineer, 
and  the  important  relations  and  methods  are  illustrated  by 
practical  numerical  problems,  of  which  there  are  several  hundred 
in  the  text.  All  matter  of  purely  historical  or  theoretical  interest 
has  been  left  out,  as  well  .as  special  topics  which  are  of  interest 
to  a  professional  designer  only.  An  ambitious  student  will  find 
a  more  exhaustive  treatment  in  the  numerous  references  given  in 
the  text. 

Many  thanks  are  due  to  the  author's  friend  and  colleague, 
Mr.  John  F.  H.  Douglas,  instructor  in  electrical  engineering  in 
Sibley  College,  who  read  the  manuscript  and  the  proofs,  checked 
the  answers  to  the  problems,  and  made  many  excellent  sugges- 
tions for  the  text.  Most  of  the  sketches  are  original,  and  are  the 
work  of  Mr.  John  T.  Williams  of  the  Department  of  Machine 
Design  of  Sibley  College,  to  whom  I  am  greatly  indebted. 

CORNELL  UNIVERSITY,  ITHACA,  N.  Y., 
September,  1911. 


CONTENTS 


PAGE 

PREFACE v 

SUGGESTIONS  TO  TEACHERS xi 

CHAPTER  I.    THE  FUNDAMENTAL  RELATION  BETWEEN  FLUX  AND  MAG- 
NETOMOTIVE FORCE 1 

A  simple  magnetic  circuit.  Magnetomotive  force.  Magnetic  flux. 
The  reluctance  of  a  magnetic  path.  The  permeance  of  a  magnetic 
path.  Reluctivity  and  permeability.  Magnetic  intensity.  Flux 
density.  Reluctances  and  permeances  in  series  and  in  parallel. 

CHAPTER  II.    THE  MAGNETIC  CIRCUIT  WITH  IRON 20 

The  difference  between  iron  and  non-magnetic  materials.  Mag- 
netization curves.  Permeability  and  saturation.  Problems  involv- 
ing the  use  of  magnetization  curves. 

CHAPTER  III.    HYSTERESIS  AND  EDDY  CURRENTS  IN  IRON 32 

The  hysteresis  loop.  An  explanation  of  saturation  and  hysteresis 
in  iron.  The  loss  of  energy  per  cycle  of  magnetization.  Eddy  cur- 
rents in  iron.  The  significance  of  iron  loss  in  electrical  machinery. 
The  total  core  loss.  Practical  data  on  hysteresis  loss.  Eddy  cur- 
rent loss  in  iron.  The  separation  of  hysteresis  from  eddy  currents. 

CHAPTER  IV.    INDUCED  E.M.F.  IN  ELECTRICAL  MACHINERY 55 

Methods  of  inducing  e.m.f.  The  formulae  for  induced  e.m.f. 
The  induced  e.m.f.  in  a  transformer.  The  induced  e.m.f.  in  an 
alternator  and  in  an  induction  motor.  The  breadth  factor.  The 
slot  factor  ks.  The  winding-pitch  factor  kw.  Non-sinusoidal  vol- 
tages. The  induced  e.m.f.  in  a  direct-current  machine.  The  ratio 
of  A.C.  to  D.C.  voltage  in  a  rotary  converter. 

CHAPTER  V.    EXCITING  AMPERE-TURNS  IN  ELECTRICAL  MACHINERY.  ...     80 

The  exciting  current  in  a  transformer.  The  exciting  current  in  a 
transformer  with  a  saturated  core.  The  types,  of  magnetic  circuit 
occurring  in  revolving  machinery.  The  air-gap  ampere-turns.  The 
method  of  equivalent  permeances  for  the  calculation  of  air-gap 
ampere-turns. 

vii 


viii  CONTENTS 

PAGE 

CHAPTER  VI.    EXCITING  AMPERE-TURNS  IN  ELECTRICAL  MACHINERY. 

(Continued) 100 

The  ampere-turns  required  for  saturated  teeth.  The  ampere- 
turns  for  the  armature  core  and  for  the  field  frame.  Magnetic 
leakage  between  field  poles.  The  permeance  and  reluctance  of  irreg- 
ular paths.  The  law  of  flux  refraction. 

CHAPTER  VII.    MAGNETOMOTIVE  FORCE  OF  DISTRIBUTED  WINDINGS  .  . .   121 

The  m.m.f.  of  a  direct-current  or  single-phase  distributed  wind- 
ing. The  m.m.f.  of  polyphase  windings.  The  m.m.fs.  in  a  loaded 
induction  machine.  The  higher  harmonics  of  the  m.m.fs. 

CHAPTER  VIII.    ARMATURE  REACTION  IN  SYNCHRONOUS  MACHINES.  .  .  .   139 

Armature  reaction  and  armature  reactance  in  a  synchronous 
machine.  The  performance  diagram  of  a  synchronous  machine 
with  non-salient  poles.  The  direct  and  transverse  armature  reac- 
tion in  a  synchronous  machine  with  salient  poles.  The  Blondel 
performance  diagram  of  a  synchronous  machine  with  salient  poles. 
The  calculation  of  the  value  of  the  coefficient  of  direct  reaction.  The 
calculation  of  the  value  of  the  coefficient  of  transverse  reaction. 

CHAPTER  IX.    ARMATURE  REACTION  IN  DIRECT-CURRENT  MACHINES  ....   163 

The  direct  and  transverse  armature  reactions.  The  calculation 
of  the  field  ampere-turns  in  a  direct-current  machine  under  load. 
Commutating  poles  and  compensating  windings.  Armature  reac- 
tion in  a  rotary  converter. 

CHAPTER  X.    ELECTROMAGNETIC  ENERGY  AND  INDUCTANCE 177 

The  energy  stored  in  an  electromagnetic  field.  Electromagnetic 
energy  expressed  through  the  linkages  of  current  and  flux.  Induc- 
tance as  the  coefficient  of  stored  energy,  or  the  electrical  inertia  of  a 
circuit. 

CHAPTER  XI.    THE  INDUCTANCE  OF  CABLES  AND  OP  TRANSMISSION 

LINES 189 

The  inductance  of  a  single-phase  concentric  cable.  The  mag- 
netic field  created  by  a  loop  of  two  parallel  wires.  The  inductance 
of  a  single-phase  line.  The  inductance  of  a  three-phase  line  with 
symmetrical  and  semi-symmetrical  spacing.  The  equivalent  reac- 
tance and  resistance  of  a  three-phase  line  with  an  unequal  spacing  of 
the  wires. 

CHAPTER  XII.    THE  INDUCTANCE  OF  THE  WINDINGS  OF  ELECTRICAL 

MACHINERY 208 

The  inductance  of  transformer  windings.  The  equivalent  leak- 
age permeance  of  armature  windings.  The  leakage  reactance  in 
induction  machines.  The  leakage  reactance  in  synchronous  machines. 
The  reactance  voltage  of  coils  undergoing  commutation 


CONTENTS  ix 

PAOE 

CHAPTER  XIII.    THE  MECHANICAL  FORCE  AND  TORQUE  DUE  TO  ELEC- 
TROMAGNETIC ENERGY 240 

The  density  of  energy  in  a  magnetic  field.  The  longitudinal 
tension  and  the  lateral  compression  in  a  magnetic  field.  The  deter- 
mination of  the  mechanical  forces  by  means  of  the  principle  of  virtual 
displacements.  The  torque  in  generators  and  motors. 

APPENDIX  1 262 

APPENDIX  II.. . .  .  266 


SUGGESTIONS  TO  TEACHERS 


(1)  THIS  book  is  intended  to  be  used  as  a  text  in  a  course 
which   comprises  lectures,   recitations,   computing  periods,   and 
home  work.     Purely  descriptive  matter  has  been  omitted  or  only 
suggested,  in  order  to  allow  the  teacher  more  freedom  in  his 
lectures  and  to  permit  him  to  establish  his  own  point  of  view. 
Some  parts  of  the  book  are  more  suitable  for  recitations,  others 
as  reference  in  the  computing  room,  others,  again,  as  a  basis  for 
discussion  in  lectures  or  for  brief  theses. 

(2)  Different  parts  of  the  book  are  made  as  much  as  possible 
independent  of  one  another,   so  that  the  teacher  can  schedule 
them  as  it  suits  him  best.     Moreover,  most  of  the  chapters  are 
written  according  to  the  concentric  method,  so  that  it  is  not 
necessary  to  finish  one  chapter  before  starting  on  the  next.     One 
can  thus  cover  the  subject  in  an  abridged  manner,  omitting  the 
last  parts  of  the  chapters. 

(3)  The  problems  given  at  the  end  of  nearly  every  article  are 
an  integral  part  of  the  book,  and  should,  under  no  circumstances, 
be  omitted.     There  is  no  royal  way  of  obtaining  a  clear  under- 
standing of  the  underlying  physical  principles,  and  of  acquiring 
an  assurance  in  their  practical  application,  except  by  the  solution 
of  numerical  examples.     It  is  convenient  to  assign  each  student 
the  complete  specifications  of  a  machine  of  each  kind,  and  ask 
him  to  solve  the  various  problems  in  the  text  in  application  to 
these  machines,  in  proportion  as  the  book  is  covered.    Numer- 
ous specifications  and  drawings  of  electrical  machines  will  be 
found  in   the  standard  works   of    E.  Arnold,  H.   M.  Hobart, 
Pichelmayer    and    others,   mentioned    in   the  footnotes  in  the 
text.     A  first-hand  acquaintance  with  these  classical  works  on 
the  part  of  the  student  is  very  desirable,  however  superficial  this 
acquaintance  may  be. 

xi 


Xif  SUGGESTIONS  TO  TEACHERS 

(4)  The  book  contains  comparatively  few  sketches;  this  gives 
the  student  an  opportunity  to  illustrate  the  important  relations 
by  sketches  of  his  own.   Making  sketches  and  drawings  of  electric 
machines  to  scale,  with  their  mechanical  features,  should  be  one 
of  the  important  features  of  an  advanced  course,  even  though  it 
may  not  be  popular  with  some  analytically-inclined  students. 
Mechanical  drawing  develops  precision  of  judgment,  and  gives 
the  student  a  knowledge  of  machinery  and  apparatus  that  is 
tangible  and  concrete. 

(5)  The  author  has  avoided  giving  definite  numerical  data, 
coefficients  and  standards,  except  in  problems,  where  they  are 
indispensable   and  where  no  general  significance  is  ascribed  to 
such  data.     His  reasons  are:  (a)  Numerical  coefficients  obscure 
the  general  exposition.    (6)  Sufficient  numerical  coefficients  and 
design  data  will  be  found  in  good  electrical  hand-books  and  pocket- 
books,  one  of  which  ought  to  be  used  in  conjunction  with  this 
text,    (c)  The  student  is  liable  to  ascribe  too  much  authority  to 
a  numerical  value  given  in  a  text-book,  while  in  reality  many 
coefficients  vary  within  wide  limits,  according  to  the  conditions 
of  a  practical  problem  and  with  the  progress  of  the  art.     (d) 
Most  numerical  coefficients  are  obtained  in  practice  by  assuming 
that  the  phenomenon  in  question  occurs  according  to  a  definite  law, 
and  by  substituting  the  available  experimental  data  into  the  corre- 
sponding formula.     This  point  of  view  is  emphasized  throughout 
the  book,  and  gives  the  student  the  comforting  feeling  that  he 
will  be  able  to  obtain  the  necessary  numerical  constants  when 
confronted  by  a  definite  practical  situation. 

(6)  The  treatment  of  the  magnetic  circuit  is  made  as  much 
as  possible  analogous  to  that  of  the  electrodyamic  and  electro- 
static circuits  treated  in  the  companion  book.     The  teacher  will 
find  it  advisable  to  make  his  students  perfectly  fluent  in  the  use 
of  Ohm's  law  for  ordinary  electric  circuits  before  starting  on  the 
magnetic  circuit.     The  student  should  solve  several  numerical 
examples  involving  voltages  and  voltage  gradients,  currents  and 
current    densities,    resistances,    resistivities,    conductances,    and 
conductivities.     He  will  then  find  very  little  difficulty  in  master- 
ing the  electrostatic  circuit,  and  with  these  two  the  transition 
to  the  magnetic  circuit  is  very  simple  indeed.     The  following 
table  shows  the  analogous  quantities  in  the  three  kinds  of  cir- 
cuits. 


SUGGESTIONS   TO  TEACHERS 


Xiii 


Electrodynamic, 

r  Voltage  or  e.m.f. 
|  Voltage  gradient  (or 
I     electric  intensity) 

Electric  current 
Current  density 

r  Resistor 
j  Resistance 
<•  Resistivity 

r  Conductor 
•j  Conductance 
I  Conductivity 


Electrostatic, 

Voltage  or  e.m.f. 
Voltage  gradient  (or 
electric  intensity) 

Dielectric  flux 
Dielectric  flux  density 

Elastor 

Elastance 

Elastivity 

Permittor  (condenser) 
Permittance  (capacity) 
Permittivity  (dielec- 
tric constant) 


Magnetic. 

Magnetomotive  force 
M.m.f.    gradient      (or 
magnetic  intensity) 

Magnetic  flux 
Magnetic  flux  density 

Reluctor 

Reluctance 

Reluctivity 

Permeator 
Permeance 
Permeability 


LIST   OF   PRINCIPAL  SYMBOLS 


The  following  list  comprises  most  of  the  symbols  used  in  the  text.  Those 
not  occurring  here  are  explained  where  they  appear.  When,  also,  a  symbol 
has  a  use  different  from  that  stated  below,  the  correct  meaning  is  given 
where  the  symbol  occurs. 


Symbol.  Meaning.                                                                                              . 

a  Air-gap  ...................................................  90 

a  Width  of  commutator  segment  ..............................  237 

A  Area  ..................................................  .  .     11 

Aa  Area  of  flux  per  tooth  pitch  in  the  air  ........................   101 

Ai  Area  of  flux  per  tooth  pitch  in  the  iron  ......................   101 

(AC)  Number  of  ampere-conductors  per  centimeter  ................   165 

b  Thickness  of  transformer  coil  ..............................   211 

b  Width  of  brush  ...........................................   236 

b'  Thickness  of  mica  ____  ,  ....................................  236 

B  Flux  density  ..............................................     14 

Bm  Maximum  value  of  the  flux  density  ..........................     81 

C2  Number  of  secondary  conductors  ...........................    133 

Cpp  Conductors  per  pole  per  phase  ..............................  219 

d  Duct  width  ..............................................     94 

e,  E  Electromotive  force  ......  ................................  39,  65 

/  Frequency  ...............................................     48 

F  Mechanical  force  ..........................................   274 

Ft  Tension  per  square  centimeter  ..............................  243 

FC  Compression  per  square  centimeter  ..........................  244 

H  Magnetic  intensity  ....................  ....................     13 

Hm  Maximum  value  of  the  magnetic  intensity  ....................     81 

i,  I  Electric  current  .......................................   39,  205 

z'o  Magnetizing  current  ............................  ...........     81 

/i  Current  per  armature  branch  ...............................  233 

k,  k'  Transformer  constant  .................................  215,  221 

ka  Air-gap  factor  ..........................  ........  ..........     90 

kb  Breadth  factor  ............................................     65 

ks  Slot  factor  ..............................  .'  ................     68 

kw  Winding-pitch  factor  ......................................     68 

I  Length  ...................................................   11 

xv 


xvi  LIST  OF   PRINCIPAL  SYMBOLS 

Symbol.  Meaning.  I*.  wb«dg,«»l 

la,  lg  Gross  armature  length 90,  102 

li  Semi-net  armature  length 220 

ln  Net  armature  length 102 

L  Inductance 184 

m  Number  of  phases 69 

M  Magnetomotive  force 7 

Ma  M.m.f.  of  armature 144 

Ma  M.m.f.  of  direct  reaction 153 

Mf  M.m.f.  of  field 144 

Mn  Net  m.m.f 144 

Mt  M.m.f.  of  transverse  reaction 153 

MI  Demagnetizing  m.m.f 165 

M 2  Distorting  m.m.f 166 

n  Number  of  turns 127,  211 

TOI,  NI  Number  of  turns  in  primary  of  a  transformer 62,  81 

N2  Number  of  turns  in  secondary  of  a  transformer 62 

N  Total  turns  in  series 65 

Om  Mean  length  of  turn 211 

p  Number  of  poles 133 

P  Power 48 

(P  Permeance 9 

(Pa  Permeance  of  air-gap 89 

(Pa  Equivalent  permeance  around  conductors  in  the  ducts  per  cm. .  220 

(Pc  Permeance  of  the  path  of  the  complete  linkages 182 

(Pe  Equivalent  permeance  around  the  end-connections  per  cm 220 

(Peg  Equivalent  permeance 184 

(Pi'  Equivalent  permeance  around  embedded  conductors  per  cm. . .  220 

(Pp  Permeance  of  the  path  of  the  partial  linkages 182 

(Ps  Permeance  of  simplified  air-gap .  . 90 

(Pz  Zig-zag  permeance 223 

q  Number  of  sections  in  a  transformer 213 

q  Number  of  turns  per  commutator  segment 235 

r,  R  Resistance 51,  177 

(R.P.M.)  Speed  in  revolutions  per  minute 260 

(R  Reluctance 7 

s  Distance 244 

s  Number  of  coils  short-circuited  by  a  brush 236 

s  Slot  width 93 

S  Number  of  slots  per  pole  per  phase 68 

S  Surface  area 244 

t  Thickness  of  laminations 51 

t  Time : 9 

t  Tooth  width 93 

t'  Tooth  width  corrected 93 

T  Time  of  one  cycle 64 

T  Torque 253 


LIST  OF  PRINCIPAL  SYMBOLS  xvii 

Symbol.  Meaning.                                                                  Pa^^«edjfined 

v            Velocity 59 

v             Volts  per  ampere  turn 155 

V           Volume 39 

w,  wp     Pole  width 90,  166 

W          Energy 39 

W         Density  of  energy 240 

Wm        Mechanical  work  done 250 

W8         Energy  stored  in  the  magnetic  field 250 

x            Reactance 144 

a           Angle 69 

a           Coefficient 220 

P            Phase  angle 150 

T            Angle 75 

d            Brush  shift  in  cm 163 

£            Eddy-current  constant '. 51 

£            Winding  pitch 71 

Ty             Hysteresis  coefficient 48 

0            Angle 253 

A            Tooth  pitch 93 

/*            Permeability 11 

v            Reluctivity 11 

T            Pole  pitch 64 

<f)           External  phase  angle 144 

f/V          Internal  phase  angle 144 

0           Flux 7 

$m         Maximum  value  of  the  flux 81 

@p         Flux  in  the  path  of  the  partial  linkages 182 

$1          Primary  leakage  flux 86 

$3          Secondary  leakage  flux 86 

X            Form  factor 65 

Amplitude  factor 84 

Angle  between  current  and  no-load  e.m.f 144 

Angle 197 


THE  MAGNETIC  CIRCUIT 


CHAPTER  I 

THE  FUNDAMENTAL  RELATION  BETWEEN  FLUX 
AND  MAGNETOMOTIVE  FORCE 

1.  A  Simple  Magnetic  Circuit.  The  only  known  cause  of 
magnetic  phenomena  is  an  electric  current,  or,  more  generally, 
electricity  in  motion.1  The  fundamental  relation  between  an 
electric  current  and  magnetism  can  be  best  studied  with  the  simple 
arrangement  shown  in  Fig.  1.  A  coil,  CC,  of  very  thin  wire  is  uni- 
formly wound  in  one  layer  on  a  spool  made  in  the  shape  of  a  circu- 
lar ring  (toroid) .  The  tubular  space  inside  of  the  ring  is  filled  with 
some  "  non-magnetic  "  material,  so  called;  for  instance,  air,  wood, 
etc.  When  a  direct  current  is  sent  through  the  coil,  the  space 
inside  the  coil  is  found  to  be  in  a  peculiar  state,  called  the  magnetic 
state.  This  magnetic  state  can  be  experimentally  proved  by 
various  means,  such  as  a  compass  needle,  iron  filings,  etc.  A 
region  in  which  a  magnetic  state  is  manifested  is  called  a  magnetic 
field.  Thus,  in  Fig.  1,  the  tubular  space  inside  the  coil  is  the  mag- 
netic field  excited  by  the  current  in  the  coil  CC. 

No  magnetic  field  is  found  in  the  space  outside  the  coil  upon 
exploring  it  with  a  magnetic  needle  or  with  iron  filings.  For 
reasons  of  symmetry,  the  field  inside  the  ring  is  the  same  at  all 
the  cross-sections.  Thus,  a  uniformly  wound  ring  constitutes 

1  Werner  v.  Siemens,  Wiedemann's  Annalen,  Vol.  24.  (1885),  p.  94;  Lar- 
mor,  Ether  and  Matter  (1904),  p.  108.  The  magnetism  of  a  permanent  magnet 
is  probably  due  to  molecular  currents  produced  by  some  orbital  motion  of 
electrons  within  the  atoms  of  iron.  The  older  concepts  of  magnetic  charges 
and  free  poles  are  summarily  dismissed  in  this  book  as  inadequate  and 
artificial. 


THE  MAGNETIC  CIRCUIT 


[ART.  1 


the  simplest  magnetic  circuit,  because  the  field  is  uniform  and  is 
entirely  confined  within  the  winding. 

Iron  filings  orient  themselves  within  the  coil  in  directions  indi- 
cated in  Fig.  I  by  the  concentric  lines  with  arrow-heads.  These 
lines  show  that  the  medium  is  "  magnetized  "  along  circles  con- 
centric with  the  ring.  Lines  which  show  the  directions  in  which  a 
medium  is  magnetized  are  generally  called  magnetic  lines  of  force.1 
They  are  analogous  to  the  lines  of  electrostatic  displacement, 
though  their  directions  and  physical  nature  are  entirely  different ; 


SECTION  A-A  C 

FIG.  1. — A  simple  magnetic  circuit. 

see   the   chapter   on   the  electrostatic    circuit,    in   the    author's 
Electric  Circuit. 

The  positive  direction  of  the  lines  of  force  is  purely  conven- 
tional, and  is  defined  as  that  in  which  the  north-seeking  end  of  a 
compass  moves.  Its  relation  to  the  current  is,  by  experiment,  that 
given  by  the  right-hand  screw  rule.  Namely,  if  the  direction  of 
the  flow  of  a  current  is  that  of  the  rotation  of  a  right-hand  screw, 
the  lines  of  force  point  in  the  direction  of  the  progressive  move- 
ment of  the  screw.  Reversing  the  current  reverses  the  direction 

1  For  actual  photographs,  showing  iron  filings  which  map  out  the  magnetic 
field  inside  of  coils  of  various  shapes,  see  Dr.  Benischke,  Die  Wissenschaft- 
lichen  Grundlagen  der  Elektrotechnik  (1907),  p.  126;  also  his  Transformatoren 
(1909),  pp.  4,  6,  and  57. 


CHAP.  I]  FLUX  AND  MAGNETOMOTIVE  FORCE  3 

of  the  field;  this  fact  can  be  demonstrated  by  a  small  compass 
needle.  The  positive  direction  of  the  lines  of  force  is  indicated 
in  Fig.  1  by  arrow  heads.  The  direction  of  the  current  is  shown 
in  the  conventional  way  by  dots  and  crosses;  namely,  a  dot 
indicates  that  the  current  is  approaching  the  observer,  while  a 
cross  indicates  that  the  current  is  receding. 

The  magnetic  state  within  the  coil  can  also  be  explored  by  a 
small  test-coil  inserted  into  the  field  and  connected  to  a  galvanom- 
eter. When  this  coil  is  properly  placed  with  respect  to  the  field 
and  then  turned  about  its  axis  by  some  angle,,  the  galvanometer 
shows  a  deflection,  because  a  current  is  induced  in  the  coil  by  the 
magnetic  field.  There  are  also  other  means  for  detecting  a  mag- 
netic field,  for  which  the  reader  is  referred  to  books  on  physics. 

The  total  magnetic  field  produced  by  a  current  is  called  a 
magnetic  circuit,  by  analogy  with  the  electric  and  the  electrostatic 
circuits.  Experiment  shows  that  the  magnetic  lines  of  force  are 
always  closed  curves  like  the  stream  lines  of  an  electric  current,  or 
like  the  lines  of  electrostatic  displacement  (when  these  are  com- 
pleted through  the  conductors). 

Fig.  1  exemplifies  a  fundamental  law  of  electromagnetism ; 
namely,  an  electric  current  creates  a  magnetic  field  in  such  direc- 
tions that  the  lines  of  force  are  linked  with  the  lines  of  flow  of  the 
current,  in  the  same  manner  that  the  consecutive  links  of  a  chain 
are  linked  together.  This  law  admits  of  no  theoretical  proof,  and 
must  be  accepted  as  a  fundamental  experimental  fact.  Wherever 
there  is  an  electric  circuit  there  is  also  a  magnetic  field  linking 
with  it.  The  two  are  inseparable,  and  increase  and  decrease 
together.  Each  form  of  an  electric  circuit  with  a  certain  strength 
of  current  in  it  corresponds  to  a  definite  form  of  magnetic  field. 
It  is  possible  that  the  electric  current  and  the  magnetic  field 
are  but  two  different  ways  of  looking  upon  one  and  the  same 
phenomenon. 

The  linkages  of  magnetic  lines  with  a  current  are  seen  more 
clearly  in  Fig.  11,  which  shows  the  magnetic  field  produced  by  a 
loop  of  wire,  aa.  It  will  be  seen  that  the  arrangement  in  Fig.  1 
is  more  suitable  for  an  elementary  study,  because  the  field  is  much 
more  uniform,  especially  if  the  radial  thickness  of  the  ring  is 
small  as  compared  to  its  mean  diameter,  so  that  all  the  lines  of 
force  are  of  practically  the  same  length. 

The  same  right-hand  screw  rule  applies  in  the  case  of  Fig.-  11 


4  THE  MAGNETIC  CIRCUIT  [ART.  2 

as  in  Fig.  1.  When  the  current  in  the  loop  of  wire  circulates  in 
the  direction  of  rotation  of  a  right-hand  screw  (toward  the  reader 
on  the  left),  the  lines  of  force  within  the  loop  point  in  the  direc- 
tion of  the  progressive  movement  of  the  screw  (upward) .  The 
rule  can  be  reversed  by  saying  that  when  the  direction  of  the 
lines  of  force  around  a  wire  is  that  of  the  rotation  of  a  right-hand 
screw,  the  current  in  the  wire  flows  in  the  direction  of  the  pro- 
gressive movement  of  the  screw.  The  first  statement  is  con- 
venient in  the  case  of  a  ring  winding,  the  second  in  the  case  of  a 
long  straight  conductor.  Both  rules  can  be  combined  into  one 
by  considering  the  exciting  electric  circuit  and  the  resulting  mag- 
netic circuit  as  two  consecutive  links  of  a  chain.  When  the  arrow- 
head in  one  of  the  links  (no  matter  which)  points  in  the  direction 
of  rotation  of  a  right-hand  screw,  the  arrow-head  in  the  other  link, 
as  it  passes  through  the  first,  must  point  in  the  direction  of  the 
progressive  movement  of  the  screw. 

2.  Magnetomotive  Force.  Experiment  shows  that  the  mag- 
netic field  within  the  ring  (Fig.  1)  does  not  change  if  the  current 
and  the  number  of  turns  of  the  "  exciting  "  winding  vary  so  that 
their  product  remains  the  same.  That  is  to  say,  500  turns  of 
wire  with  a  current  of  2  amperes  flowing  through  each  will  pro- 
duce the  same  field  as  1000  turns  with  1  ampere,  or  200  turns  with 

5  amperes,  because  the  product  is  equal  to  1000  ampere-turns  in 
all  cases.     Even  one  turn  with  1000  amperes  flowing  through  it 
will  produce  the  same  effect,  provided  that  the  turn  is  made  of 
a  wide  sheet  of  metal  spread  over  the  whole  surface  of  the  ring, 
so  as  to  make  its  action  uniform  throughout. 

The  reason  for  the  above  can  be  seen  by  considering  1000 
separate  turns  with  a  current  of  1  ampere  flowing  through  each 
turn,  and  each  turn  supplied  with  current  from  an  independent 
electrical  source,  say  a  dry  cell.  Connecting  all  the  cells  and  all 
the  turns  in  series  gives  1000  turns  with  one  ampere  flowing 
through  each.  Connecting  the  cells  and  the  turns  in  parallel 
results  in  one  wide  turn  with  1000  amperes  of  current  in  it. 
Such  changes  in  the  electrical  connections  cannot  affect  the  action 
of  each  current  outside  the  wire,  because  the  value  of  the  current 
and  the  position  of  the  turn  is  the  same  in  both  cases.  Hence, 
the  magnetic  action  depends  only  upon  the  number  of  turns  each 
carrying  1  ampere,  in  other  words,  it  depends  upon  the  num- 
ber of  ampere-turns. 


CHAP.  I]  FLUX  AND  MAGNETOMOTIVE  FORCE  5 

The  number  of  ampere-turns  of  the  exciting  winding  is  called 
the  magnetomotive  force  of  the  magnetic  circuit,  because  these 
ampere-turns  are  the  cause  of  the  magnetic  field.  One  ampere- 
turn  is  the  logical  unit  of  magnetomotive  force.  In  the  example 
above,  the  magnetomotive  force  is  equal  to  1000  ampere-turns. 
In  electric  machines  the  field  excitation  often  reaches  several 
thousand  ampere-turns,  and  the  magnetomotive  force  is  for  con- 
venience sometimes  measured  in  kiloampere-turns,  one  kilo- 
ampere-turn  being  equal  to  1000  ampere-turns. 

3.  Magnetic  Flux.  The  magnetic  disturbance  at  each  point 
within  the  ring  has  not  only  a  direction,  but  also  a  magnitude.  The 
disturbance  is  said  to  be  in  the  form  of  a  flux,  for  the  following 
reason :  One  may  think  of  the  magnetic  state  as  being  due  to  the 
actual  displacement  of  some  hypothetical  incompressible  sub- 
stance along  the  lines  of  force;  in  this  case  the  flux  represents  the 
amount  of  this  substance  displaced  through  each  cross-section  of 
the  ring,  and  is  analogous  to  total  electrostatic  displacement.  Or, 
as  some  modern  writers  think,  there  is  an  actual  flow  of  an  incom- 
pressible ether  along  the  lines  of  force.  In  that  case  the  flux  may 
be  thought  of  as  the  rate  of  flow  of  the  ether  through  a  cross-sec- 
tion. The  viewpoint  common  to  these  two  explanations  gave 
rise  to  the  name  flux  which  means  flow. 

Some  physicists  consider  the  magnetic  circuit  as  consisting 
of  infinitely  subdivided  (though  closed)  whirls  or  vortices  in  the 
ether,  the  rotation  being  in  planes  perpendicular  to  the  lines  of 
force.  Each  line  of  force  is  considered,  then,  as  the  geometric 
axis  of  an  infinitely  thin  fiber  or  tube  of  force,  and  the  ether  within 
each  tube  in  a  state  of  transverse  vortex  motion.  The  line  of 
force  represents  the  direction  of  the  axis  of  rotation,  and  the  flux 
may  be  thought  of  as  the  momentum  of  the  rotating  substance 
per  unit  length  of  the  tubes  of  force.  According  to  any  of  these 
three  views,  the  energy  of  a  current  is  actually  contained  in  the 
magnetic  circuit  linked  with  the  current. 

Whichever  view  is  adopted,  the  magnetic  flux  can  be  defined  as 
the  sum  total  of  magnetic  disturbance  through  a  cross-section  per- 
pendicular to  the  lines  of  force.  Experiment  shows  that  the  total 
flux  is  the  same  through  all  complete  cross-sections  of  a  magnetic 
circuit.  This  could  have  been  expected  from  the  point  of  view 
of  a  displacement  or  flow  along  the  lines  of  force;  each  tube  of 
force  being  like  a  channel  within  which  the  displacement  or  the 


6  THE  MAGNETIC  CIRCUIT  [ART.  3 

flow  of  an  incompressible  substance  takes  place.  For  this  reason 
the  magnetic  flux  is  said  to  be  solenoidal  (i.e.,  channel-shaped). 

The  familiar  law  of  electromagnetic  induction  discovered  by 
Faraday  is  used  for  the  definition  of  the  unit  of  flux.  Namely,  when 
the  total  magnetic  disturbance  or  flux  within  a  turn  of  wire 
changes,  an  electromotive  force  is  induced  in  the  turn.  By  experi- 
ments in  a  uniform  field,  the  fact  is  established  that  the  value  of 
the  induced  electromotive  force  is  exactly  proportional  to  the  rate 
of  change  of  the  flux  linking  with  the  test  loop.  This  fact  is 
used  in  the  definition  of  the  unit  of  flux. 

With  the  volt  and  the  second  as  the  units  of  e.m.f.  and  of 
time  respectively,  the  corresponding  unit  of  flux  is  called  the 
weber }  and  is  defined  as  follows :  A  flux  through  a  turn  of  wire 
changes  at  a  uniform  rate  of  one  weber  per  second  when  the  e.m.f. 
induced  in  the  turn  remains  constant  and  equal  to  one  volt.  Such 
a  unit  flux  can  be  also  properly  called  the  volt-second,  though  as 
yet  neither  name  has  been  recognized  by  the  International  Electro- 
technical  Commission.  The  weber  or  the  volt-second  is  too  large 
a  unit  for  most  practical  purposes.  Therefore  a  much  smaller 
unit,  called  the  maxwell,1  is  used,  which  is  equal  to  one  one-hun- 
dred-millionth part  of  the  weber,  or 

one  maxwell  =  one  weber  XlO~8. 

The  lines  of  force  in  Figs.  1  and  11  can  be  made  to  represent 
not  only  the  direction  of  the  field,  but  its  magnitude  as  well,  if  they 
be  drawn  at  suitable  distances  from  each  other.  That  is,  such 
that  the  total  number  of  lines  passing  through  any  part  of  a 
cross-section  of  the  ring  is  equal  numerically  to  the  number  of 
maxwells  in  the  flux  through  the  same  part.  With  this  conven- 
tion, each  line  stands  symbolically  for  one  maxwell;  some  engi- 
neers and  physicists  speak  of  the  number  of  lines  of  force  in  a  flux 
when  they  mean  maxwells. 

While  the  weber  is  too  large  a  unit,  the  maxwell  is  too  small  for 
many  practical  purposes.  Therefore  two  other  intermediate  units 

1  The  origin  of  the  maxwell  becomes  clear  when  one  remembers  that  the 
volt  was  originally  established  as  108  electromagnetic  C.G.S.  unit  of  electro- 
motive force.  The  maxwell  is  related  to  the  C.G.S.  unit  of  e.m.f.  or  the  so- 
called  abvolt  in  the  same  way  in  which  the  weber  is  related  to  the  ordinary 
volt.  In  other  words,  when  the  flux  within  a  coil  varies  at  the  rate  of  cne 
maxwell  per  second,  one  abvolt  is  induced  in  each  turn  of  the  winding. 


CHAP.  I]  FLUX  AND  MAGNETOMOTIVE  FORCE  7 

are  used,  namely  the  kilo-maxwell,  equal  to  one  thousand  max- 
wells, and  the  mega-maxwell,  equal  to  one  million  maxwells. 
These  two  units  are  sometimes  called  the  kilo-line  and  the  mega- 
line,  the  word  "  line  "  being  used  for  the  word  maxwell,  as 
explained  above.1 

Prob.  1.  The  flux  within  the  coil  (Fig.  1)  is  equal  to  63  kilo-maxwells. 
A  test  coil  of  five  turns  is  wound  on  the  exciting  coil  so  as  to  be  linked 
with  the  total  flux.  What  voltage  is  induced  in  this  test  coil  when  the 
current  in  the  main  (exciting)  coil  is  reduced  to  zero  at  a  uniform  rate 
in  seven  seconds?  Ans.  0.45  millivolt. 

Prob.  2.  At  what  rate  must  the  flux  be  varied  in  the  preceding 
problem  in  order  to  induce  one  volt  in  the  test  coil? 

Ans.  '  0.2  weber  (20  megalines)  per  second. 

Prob.  3.  When  the  flux  varies  at  a  non-uniform  rate  show  that  the 
voltage  induced  in  the  test  coil  at  any  instant  is  equal  to  (d<P  /dt)  X  10-, 
where  t  is  time  in  seconds,  and  0  is  the  flux  in  maxwells.  Show  that 
the  exponent  of  10  must  be  —2  instead  of  —8  if  the  flux  is  expressed  in 
megalines. 

4.  The  Reluctance  of  a  Magnetic  Path.  Experiment  shows  that 
the  total  flux  within  the  coil  (Fig.  1)  is  proportional  to  the  applied 
magnetomotive  force,  when  the  space  inside  is  filled  with  air. 
Therefore,  a  relation  similar  to  Ohm's  law  holds,  namely, 

M  =  (R0,      ........     (1) 


where  M  is  the  magnetomotive  force  in  ampere-turns,  $  is  the  flux 
in  maxwells,  and  (ft  is  the  coefficient  of  proportionality  between 
the  two,  called  the  reluctance  of  the  magnetic  circuit.  Script  (ft  is 
used  to  distinguish  reluctance  from  electric  resistance.  The  mag- 
netomotive force  M  is  the  cause  of  the  flux;  or,  with  .  reference  to 
an  electric  circuit,  M  is  analogous  to  the  applied  electromotive 
force.  0  is  analogous  to  the  resulting  current,  and  the  reluctance 
(ft  takes  place  of  the  electric  resistance.  Therefore,  eq.  (1)  is 
known  as  Ohm's  law  for  the  magnetic  circuit.  Of  course,  the 

1  This  possibility  of  creating  new  units  of  convenient  size  is  a  great 
advantage  of  the  metric  or  decimal  system  of  units.  New  units  are  gener- 
ally understood,  by  the  use  of  Latin  and  Greek  prefixes,  signifying  their 
numerical  relation  to  the  fundamental  unit.  For  instance,  it  is  perfectly 
legitimate  to  use  such  units  as  deci-ampere  and  hecto-volt,  in  spite  of  the 
fact  that  they  are  not  in  general  use.  Anyone  familiar  with  the  agreed 
prefixes  will  know  that  the  units  spoken  of  are  equal  to  one-tenth  of  one 
ampere,  and  to  one  hundred  volts.  See  Appendix  I  on  the  Ampere-Ohm 
System. 


8  THE  MAGNETIC  CIRCUIT  [ART.  4 

analogy  is  purely  formal,  the  two  sets  of  phenomena  being  entirely 
different.  An  equation  similar  to  (1)  can  be  written  for  the  flow 
of  heat,  of  water,  etc.  It  merely  expresses  the  experimental  fact 
that,  for  a  certain  class  of  phenomena,  the  effect  is  proportional 
to  the  cause. 

If  the  space  within  the  coil  be  filled  with  practically  any  known 
substance,  solid,  liquid,  or  gaseous,  the  reluctance  (R  remains 
within  less  than  ±  1  per  cent  of  the  value  which  obtains  with  air. 
The  notable  exceptions  are  iron,  cobalt,  nickel,  manganese,  chro- 
mium, and  some  of  their  oxides  and  alloys.1  When  the  circuit 
includes  one  of  these  so-called  "  ferro-magnetic  "  substances,  a 
much  larger  flux  is  produced  with  the  same  m.m.f.,  that  is,  the 
reluctance  of  the  circuit  is  apparently  reduced  to  a  considerable 
extent.  Moreover,  this  reluctance  is  no  longer  constant,  but 
depends  upon  the  value  of  the  flux.  The  behavior  of  iron  and 
steel  in  a  magnetic  circuit  is  of  great  practical  importance,  and  is 
treated  in  detail  in  Chapters  II  and  III. 

The  definition  of  the  unit  of  reluctance  follows  directly  from 
eq.  (1).  A  magnetic  circuit  has  a  unit  reluctance  when  a  magneto- 
motive force  of  one  ampere-turn  produces  in  it  a  flux  of  one 
maxwell.2  No  name  has  been  given  to  this  unit  so  far.  The  author 
ventures  to  suggest  the  name  rel,  and  he  uses  it  provisionally  in  this 
book.  Granting  that  reluctance  is  a  useful  quantity  in  magnetic 
calculations,  one  must  admit  that  it  should  be  measured  in  some 
units  of  its  own ;  unless  one  chooses  to  use  the  cumbersome  nota- 
tion "  ampere-turns  per  maxwell."  The  name  rel  is  simply  the 
beginning  of  the  word  reluctance.  Thus,  a  magnetic  circuit  has 
a  reluctance  of  one  rel  when  one  ampere-turn  produces  one 
maxwell  of  flux  in  it.  Tha  unit  rel  is  analogous  to  the  ohm  in  the 
electric  circuit,  and  to  the-ffar^|m  the  electrostatic  circuit. 

Prob.  4.  What  is  the  reluctance  of  the  magnetic  circuit  in  Fig.  1 
if  47,600  ampere-turns  produce  a  flux  of  2.3  kilo-maxwells? 

Ans.    47,600/2300  =  20.7  rels. 

Prob.  5.  How  many  ampere-turns  are  required  to  establish  a  flux 
of  1.7  megalines  through  a  reluctance  of  0.0054  rel?  Ans.  9180. 

Prob.  6.  A  wooden  ring  is  temporarily  wound  with  330  turns  of 
wire;  when  a  current  of  25  amperes  is  flowing  through  the  winding  the 

^ee  Dr.  C.P.  Steinmetz,  Magnetic  Properties  of  Material  Electrical 
World,  Vol.  55  (1910),  p.  1209. 

2  See  Appendix  I  at  the  end  of  the  book. 


CHAP.  I]  FLUX  AND  MAGNETOMOTIVE  FORCE  9 

flux  is  found  to  be  equal  to  21  kilo-maxwells.  The  permanent  winding 
on  the  same  ring  must  produce  a  flux  of  65.1  kilolines  at  a  current  of 
9.3  amperes.  How  many  turns  will  be  required?  Ans.  2750. 

5.  The  Permeance  of  a  Magnetic  Path.  In  calculations  per- 
taining to  the  electric  circuit  it  is  convenient  to  deal  with  the  recip- 
rocals of  resistances  when  conductors  are  connected  in  parallel. 
The  reciprocal  of  a  resistance  is  called  a  conductance  and  is  meas- 
ured in  mhos  if  resistance  is  measured  in  ohms.  Similarly,  a  die- 
lectric is  characterized  sometimes  by  its  elastance,  at  other  times 
by  the  reciprocal  of  its  elastance,  which  is  called  permittance. 
When  permittance  is  measured  in  farads,  elastance  is  measured  in 
darafs  (see  the  chapter  on  the  Electrostatic  Circuit  in  the  author's 
Electric  Circuit) . 

Analogously,  when  two  or  more  magnetic  paths  are  in  parallel 
it  is  convenient  to  use  the  reciprocals  of  the  reluctances.  The 
reciprocal  of  the  reluctance  of  a  magnetic  path  is  called  its  per- 
meance; eq.  (1)  becomes  then 

$  =  (PM,     .     .    / (2) 

where 

(P=1/(R (3) 

A  script  (P  is  used  for  permeance  in  order  to  avoid  confusing  it 
with  power.  For  the  unit  of  permeance  corresponding  to  the 
rel,  the  author  proposes  the  name  perm.  A  magnetic  path  has  a 
permeance  of  one  perm  when  one  maxwell  of  flux  is  produced  for 
each  ampere-turn  of  magnetomotive  force  applied  along  the  path. 

The  unit  "-perm  "  has  been  in  use  among  electrical  designers 
for  some  time,  although  no  name  has  been  given  to  it.  Notably 
Mr.  H.  M.  Hobart  has  used  it  extensively  in  his  writings, 
in  the  calculation  of  the  inductance  of  windings.  He  speaks  of 
"  magnetic  lines  per  ampere-turn  per  unit  length "  (of  the 
embedded  part  of  a  coil) .  This  is  equivalent  to  perms  per  unit 
length. 

In  the  ampere-ohm  system  the  internationally  accepted  unit 
of  permeance  is  the  henry.1  Therefore,  if  in  eq.  (2)  M  is  measured 
in  ampere-turns  and  $  in  webers,  (P  is  in  henrys,  and  no  new  unit 
for  permeance  is  necessary.  In  this  case  the  reluctance  (R  in  eqs. 

1  Although  the  henry  is  defined  as  the  unit  of  inductance,  it  is  shown  in 
Art.  58  below  that  permeance  and  inductance  are  physically  of  the  same 
dimensions  and  hence  measureable  in  the  same  units. 


10  THE  MAGNETIC  CIRCUIT  [ABT.  6 

(1)  and  (3)  is  in  henrys"1;  or  spelling  the  word  henry  backwards, 
as  in  the  case  of  mho  and  daraf,  the  natural  unit  of  reluctance  in 
the  ampere-ohm  system  gets  the  euphonious  name  of  yrneh  (to  be 
pronounced  earney). 

Since,  however,  the  maxwell  is  used  almost  exclusively  as  the 
unit  of  flux,  it  seems  advisable  to  introduce  the  rel  and  the  perm 
as  units  directly  related  to  it.  Should  engineers  gradually  feel 
inclined  to  use  the  weber  and  its  submultiples  as  the  units  of  flux, 
then  the  henry,  the  yrneh,  and  their  multiples  and  submultiples 
would  naturally  be  used  as  the  corresponding  units  of  permeance 
and  reluctance. 

We  have,  therefore,  the  two  following  systems  of  units  for 
reluctance  and  permeance,  according  to  whether  the  maxwell  or 
the  weber  is  used  for  the  unit  of  flux  (one  ampere-turn  being  the 
unit  of  m.m.f.  in  both  cases) : 


Unit  of  flux 

Unit  of  permeance 

Unit  of  reluctance 

Maxwell 
Weber 

Perm 
Henry 

Rel 
Yrneh 

L/V/A  _E— t-V/J-JLJ-    V  JLAAAV^i. 

One  perm  =  10~  8  henry ;    one  rel  =  10  8  yrnehs. 

Prob.  7.  What  is  the  permeance  of  the  magnetic  circuit  in  prob.  4? 

Ans.     0.0483  perm.  =4.83  X 10-10  henry. 
Prob.  8.   What  is  the  permeance  of  the  ring  in  prob.  6? 

Ans.     2.545  perm.  =  0.02545  microhenry. 

Prob.  9.  How  many  ampere-turns  are  required  to  maintain  a  flux 
of  2.7  megalines  through  a  permeance  of  750  perms?  Ans.  3600. 

6.  Reluctivity  and  Permeability.  The  reluctance  of  a  magnetic 
path  varies  with  the  dimensions  of  the  path  according  to  the  same 
law  as  the  resistance  of  an  electric  conductor  or  the  elastance  of  a 
dielectric.  That  is  to  say,  the  reluctance  is  directly  proportional 
to  the  average  length  of  the  lines  of  force  and  is  inversely  propor- 
tional to  the  cross-section  of  the  path.  This  relationship  can  be 
verified  by  measurements  on  rings  of  different  dimensions  (Fig.  1) . 
When  the  diameter  of  the  ring  is  increased  twice,  keeping  the  same 
cross-section,  the  length  of  the  path  of  the  flux  is  also  increased 
twice.  Experiment  shows  that  the  new  ring  requires  twice  as 
many  ampere-turns  as  the  first  one  for  the  same  flux ;  or,  only  one- 
half  of  the  flux  is  produced  with  the  same  number  of  ampere- 


CHAP.  1]  FLUX  AND  MAGNETOMOTIVE  FORCE  11 

turns.  If  the  diameter  of  the  ring  is  kept  the  same  but  the  cross- 
section  of  the  path  is  increased  twice,  the  flux  is  doubled  with  the 
same  magnetomotive  force.  These  and  similar  experiments  show 
that  the  reluctance  and  the  permeance  of  a  uniform  magnetic  path 
obey  the  same  law  as  the  resistance  and  the  conductance  of  a  con- 
ductor, or  the  elastance  and  the  permittance  of  a  prismatic  slab 
of  a  dielectric. 

We  can  therefore  put 

(ft  =  vl/A,    .     ...    .    .    .     .     .     (4) 

where  I  is  the  mean  length  of  the  path,  A  is  its  cross-section,  and 
v  is  a  physical  constant.  By  analogy  with  resistivity  and  elastiv- 
ity,  v  is  called  the  reluctivity  of  a  magnetic  medium.  If  (R  is  in  rels, 
and  the  dimensions  of  the  circuit  are  in  centimeters,  v  is  in  rels  per 
centimeter  cube.  In  other  words,  the  reluctivity  of  a  magnetic 
medium  is  the  reluctance  of  a  unit  cube  of  this  medium  when  the 
lines  of  force  are  parallel  to  one  of  the  edges.  For  air  and  all  other 
non-magnetic  substances  the  experimental  value  of  v  is  0.8  rel  per 
centimeter  cube,1  or  0.313  rel  per  inch  cube. 

The  expression  for  permeance  corresponding  to  eq.  (4)  is 


(5) 


where  the  coefficient  /*  is  called  the  permeability  of  the  magnetic 
medium.  It  corresponds  to  the  electric  conductivity  ?  and  the 
dielectric  permittivity  K.  Since  the  permeance  of  a  path  is  the 
reciprocal  of  its  reluctance,  the  permeability  of  a  medium  is  the 
reciprocal  of  its  reluctivity,  or 


(6) 


When  the  perm  and  the  centimeter  are  used  for  the  units  of  per- 
meance and  length,  permeability  is  expressed  in  perms  per  centi- 
meter cube.  For  all  non-magnetic  materials  //=1.25  perms  per 
centimeter  cube  (more  accurately  1.257).  With  the  henry  and  the 
centimeter  as  units  /*=  1.257  X10~8  henries  per  centimeter  cube. 
In  the  English  system  ^=3.19  perms  per  inch  cube  for  non- 

lMore  accurately  0.796  rel  per  centimeter  cube.  As  a  rule,  magnetic 
calculations  are  much  less  accurate  than  electrical  calculations,  because  there 
is  no  "magnetic  insulator"  known,  so  that  there  is  always  some  magnetic 
leakage  present,  which  is  difficult  to  take  into  consideration.  For  this  reason 
the  value  0.8  is  sufficiently  accurate  for  most  practical  purposes. 


12  THE  MAGNETIC    CIRCUIT  [Aux.  7 

magnetic  materials.  For  magnetic  materials  JJL  is  considerably 
larger  than  for  non-magnetic,  and  varies  with  the  field  strength. 
In  calculations  either  reluctivity  or  permeability  is  used,  according 
to  the  conditions  of  the  case  and  the  preference  of  the  engineer. 

The  student  has  probably  heard  before  that  the  permeability  of 
air  is  assumed  equal  to  unity.  The  discrepancy  between  this  com- 
monly accepted  value  and  the  value  1.25  given  above,  is  due  to  a 
different  unit  of  magnetomotive  force,  called  the  gilbert,  which  is 
sometimes  employed.  The  author  considers  the  gilbert  to  be 
of  doubtful  utility,  for  reasons  stated  in  Appendix  II ;  hence  no 
use  is  made  of  it  in  this  book. 

Prob.  10.  Assuming  the  value  of  /*  =  1.257  to  be  given,  check  the 
value  of  //  =  3.19  in  the  English  system,  and  also  the  values  of  y  in  the 
metric  and  the  English  systems,  as  given  above. 

Prob.  11.  In  prob.  4  the  reluctance  of  a  ring  was  20.7  rels.  If  the 
cross-section  of  the  ring  is  120  sq.  mm.,  what  is  the  average  diameter  of 
the  ring?  Ans.  9.9  cm. 

Prob.  12.  How  many  ampere-turns  are  required  to  establish  a  flux 
of  47  kilolines  in  a  ring  of  rectangular  cross-section,  made  of  non-magnetic 
material;  the  radial  thickness  of  the  ring  is  8  cm.,  the  axial  width  11  cm. 
and  the  average  radius  16  cm?  Ans.  About  43  kiloamp ere- turns. 

Prob.  13.  How  many  ampere-turns  would  be  required  in  the  preced- 
ing problem  for  the  same  flux  if  the  ring  were  made  of  iron,  the  relative 
permeability  of  which  (with  respect  to  air)  is  500? 

Ans.    86  ampere-turns. 

7.  Magnetic  Intensity.  In  order  that  the  student  may  better 
appreciate  the  significance  of  the  concept  of  magnetic  intensity, 
it  is  advisable  to  refresh  in  his  mind  the  corresponding  quantity 
used  in  the  electric  circuit,  viz.,  the  electric  intensity.  Namely,  in 
problems  on  the  electric  and  the  electrostatic  circuit  it  is  some- 
times desirable  to  consider  not  only  the  total  voltage,  but  also 
the  voltage  used  up  or  balanced  per  unit  length  of  the  path  along 
which  the  electricity  flows  or  is  displaced.  This  quantity,  the 
rate  of  change  of  voltage  along  the  circuit,  is  known  as  the  electric 
intensity,  or  the  voltage  gradient.  It  is  denoted  by  F  (see  the 
Electric  Circuit),  and  is  measured  in  volts  per  linear  centimeter. 
When  the  voltage  drop  is  uniform  along  a  conductor  or  a  dielectric, 
F=E/l,  where  E  is  the  voltage  between  the  ends  of  the  part  of  the 
circuit  under  consideration,  and  I  is  the  corresponding  length. 
When  the  voltage  drop  is  not  uniform,  F  is  different  for  different 
points  along  the  path,  and  for  each  point  F=dE/dl. 


CHAP.  I]  FLUX  AND  MAGNETOMOTIVE  FORCE  13 

In  a  similar  way,  the  magnetomotive  force  of  a  magnetic  circuit 
is  used  up  bit  by  bit  in  the  consecutive  parts  of  the  circuit.  One 
can  speak  not  only  of  the  total  magnetomotive  force  of  a  closed 
circuit,  but  also  of  the  magnetomotive  force  acting  upon  a  certain 
part  of  the  circuit,  and  of  magnetomotive  force  per  unit  length  of 
the  lines  of  force.  Thus,  for  instance,  if  1000  ampere-turns  is  con- 
sumed in  a  uniform  magnetic  circuit  4  cm.  long,  the  magnetomo- 
tive force  per  unit  length  of  path  is  250  ampere-turns. 

The  magnetomotive  force  per  unit  length  of  path  is  called  the  mag- 
netic intensity  at  a  point,  or  the  m.m.f.  gradient,  and  is  denoted  by 
H.  Thus,  if  the  circuit  is  uniform,  the  magnetic  intensity  at  any 
point  is 

H-M/l,        .    .     .    ....     (7) 

where  M  is  the  magnetomotive  force  acting  upon  the  length  I  of 
the  circuit.  If  the  magnetic  circuit  is  non-uniform,  for  instance,  if 
the  cross-section  of  the  ring  is  different  at  different  places,  or  if  the 
permeability  is  different  at  some  parts  of  the  circuit  due  to  the 
presence  of  iron,  the  m.m.f.  gradient  is  different  at  different  points, 
and  at  each  point  it  is  expressed  by  the  equation 

H=dM/dl,       .    ....    .......     (8) 

where  dM  is  the  m.m.f.  necessary  for  establishing  the  flux  in  the 
length  dl  of  the  circuit.     If  M  is  in  ampere-turns,  and  I  is  in  centi- 
meters, H  is  in  ampere-turns  per  centimeter. 
Eqs.  (7)  and  (8)  can  be  also  written  in  the  form 

M=m,  .........   (9) 

and 


-  C2Hdl. 

Jh 


(10) 


These  formulae,  expressed  in  words,  simply  mean  that  the  magneto- 
motive force  acting  upon  a  certain  part  of  a  magnetic  circuit  is  the 
line  integral  of  the  magnetic  intensity  along  the  path,  or  the  sum  of 
the  m.m.fs.  used  up  in  the  elementary  parts  of  the  path.  The  rela- 
tion between  M  and  H  will  become  clearer  to  the  student  in  the 
various  applications  that  follow. 

Prob.  14.  What  is  the  magnetic  intensity  in  prob.  12? 

Ans.    About  425  ampere-turns  per  cm.  of  path. 


14  THE  MAGNETIC  CIRCUIT  [ART.  8 

8.  Flux  Density.  It  is  often  of  importance  to  consider  the  flux 
density,  or  the  value  of  a  flux  per  unit  of  cross-section  perpendicular 
to  the  direction  of  the  lines  of  force.  Flux  density  is  usually 
denoted  by  B,  and  is  measured  in  maxwells  (or  its  multiples)  per 
square  centimeter.1  When  the  flux  is  distributed  uniformly  over 
the  cross-section  of  a  path,  the  flux  density 

.......     (11) 


where  A  is  the  area  of  the  cross-section  of  the  path.  If  the  flux  is 
distributed  non-uniformly,  an  infinitesimal  flux  d(D  passing  through 
a  cross-section  dA  must  be  considered.  In  the  limit,  the  flux  den- 
sity at  a  point  corresponding  to  dA  is 


(12) 


The  areas  A  and  dA  are  understood  to  be  at  all  points  normal 
to  the  direction  of  the  field.  Solving  these  two  equations  for  the 
flux  we  find 

•-B-A, (is) 

or 

,oAB-dA,  .     .    .    .    .     .     (14) 

the  integration  being  extended  over  the  whole  cross-section  of  the 
path.  Expressed  in  words,  these  last  two  formulae  mean  that 
the  total  flux  passing  through  a  surface  is  equal  to  the  sum  of  the 
fluxes  passing  through  the  different  parts  of  that  surface. 

Magnetic  flux  density  is  analogous  to  current  density  C7,  and 
to  dielectric  flux  density  D  treated  in  the  Electric  Circuit.  The 
student  will  find  no  difficulty  in  interpreting  eqs.  (11)  to  (14) 
from  the  point  of  view  of  the  electric  and  electrostatic  circuits. 

The  relation  between  B  and  H  is  obtained  from  eq.  (1)  in  which 
the  value  of  (R  is  obtained  from  eq.  (4).  Namely,  we  have 


or 


1  Some  writers  express  flux  density  in  gausses,  one  gauss  being  equal  to 
one  maxwell  per  square  centimeter.  The  unit  kilogauss,  equal  to  one  kilo- 
maxwell  per  square  centimeter,  is  also  used.  While  the  terms  gauss  and 
kilogauss  are  convenient  abbreviations,  no  use  is  made  of  them  in  this  book 
in  order  to  keep  the  relation  between  a  flux  and  the  cross-section  of  its  path 
explicitly  before  the  student. 


CHAP.  I]  FLUX  AND   MAGNETOMOTIVE  FOR   E  15 

The  last  expression,  according  to  eqs.  (7)  and  (11),  can  be  written 
simply  as 

H=Bv, (15) 

or,  since  v=l/p, 

B  =  tiH :-.'..  ,(    .     (16) 

Eqs.  (15)  and  (16)  state  Ohm's  law  for  a  unit  magnetic  path,  for 
instance,  a  path  one  centimeter  long  and  one  square  centimeter  in 
cross-section.  H  is  the  magnetomotive  force  between  the  oppo- 
site faces  of  the  cube,  fi  is  the  permeance  of  the  cube,  and  B  is  the 
flux  passing  through  it.  The  reader  will  remember  similar  equa- 
tions U=fF  and  D  =  KF  for  the  unit  electrical  conductor  and  the 
unit  prism  of  a  dielectric  respectively. 

Instead  of  beginning  the  theory  of  the  magnetic  circuit  with 
eq.  (1)  and  developing  it  into  eq.  (16),  it  is  possible  to  begin  it  with 
eq.  (16).  Namely,  the  known  magnetic  phenomena  show  that 
at  each  point  in  the  medium  there  is  a  magnetic  intensity  H  which 
is  the  cause  of  the  magnetic  state,  and  that  the  effect  is  measured 
by  the  flux  density  5;  /*  is  the  physical  constant  which  shows  the 
proportionality  between  H  and  B.  The  magnetic  circuit  is  then 
assumed  to  be  built  up  of  infinitesimal  tubes  of  flux  in  series  and 
in  parallel,  and  finally  eq.  (1)  is  obtained. 

Prob.  15.  What  is  the  flux  density  in  prob.  12? 

Ans.    534  maxwells  per  square  centimeter  (534  gausses). 

Prob.  16.  How  many  ampere-turns  per  pole  are  required  to  establish 
a  flux  density  of  7  kilolines  per  square  centimeter  in  the  air-gap  of  a 
machine,  the  clearance  being  3  mm.?  Solution:  According  to  eq.  (15) 
H  =  7000/ 1 .25  =  5600  ampere-turns  per  centimeter  of  length.  Hence  the 
required  m.m.f.  is  5600X0.3  =  1680  ampere-turns. 

9.  Reluctances  and  Permeances  in  Series  and  in  Parallel.     In 

practice,  one  has  to  deal  mostly  with  magnetic  circuits  of  irregular 
form,  for  instance,  those  of  electric  machines  (Fig.  24)  in  which 
the  flux  is  established  partly  in  air  and  partly  in  iron,  each  of  vary- 
ing cross-section.  The  circuit  consists  in  this  case  of  several  reluc- 
tances in  series.  One  may  say,  for  instance,  that  the  total  mag- 
netomotive force  required  in  this  machine,  per  magnetic  circuit, 
is  8000  ampere-turns,  of  which  6000  are  used  in  the  air-gap,  1500 
in  the  field  frame,  and  500  in  the  armature.  This  is  analogous  to 
distinguishing  between  the  total  e.m.f .  of  an  electric  circuit,  and  the 
voltage  drop  in  the  various  parts  of  the  circuit. 


16  THE  MAGNETIC  CIRCUIT  [ART.  9 

In  some  cases  two  or  more  magnetic  paths  are  in  parallel,  for 
instance,  when  there  is  magnetic  leakage  (see  below).  In  most 
cases  the  engineer  has  to  consider  complicated  magnetic  circuits 
which  consist  partly  of  paths  in  series,  partly  of  paths  in  parallel. 
Thus,  in  the  same  machine,  the  m.m.f  .  or  the  difference  of  mag- 
netic potential  between  the  pole-tips  is  6500  ampere-turns.  This 
m.m.f.  maintains  a  useful  flux  of  say  2.5  megalines  through  the 
armature,  and  say  0.5  megaline  of  leakage  or  stray  flux  between 
the  pole-tips.  Thus  the  total  flux  in  the  field  frame  is  3  mega- 
lines. 

The  fundamental  law  of  the  magnetic  circuit,  as  expressed  by 
eq.  (1),  is  analogous  to  Ohm's  law  for  the  simple  electric  circuit. 
Therefore  magnetic  paths  in  series  and  in  parallel  are  combined 
according  to  the  same  rule  that  ele'ctrical  conductors  are  combined 
in  series  and  in  parallel.  Namely,  when  two  or  more  magnetic 
paths  are  in  series,  their  reluctances  are  added  ;  when  two  or  more 
magnetic  paths  are  in  parallel  their  permeances  are  added.  Or, 
for  a  series  combination, 

........     (17) 


and  for  a  parallel  Combination 

(18) 


It  will  be  remembered  that  similar  relations  hold  also  for  impe- 
dances and  admittances  in  the  alternating  current  circuit,  and 
for  elastances  and  permittances  in  the  electrostatic  circuit. 

The  proof  of  formulae  (17)  and  (18).  is  similar  to  that  usually 
given  for  the  combination  of  electric  resistances  in  series  and  in 
parallel.  Namely,  when  reluctances  are  in  series  the  total  mag- 
netomotive force  is  equal  to  the  sum  of  component  m.m.f  .s.,  or 

.    .    .  •.-"-.    .     .     (17a) 


Dividing  both  sides  of  this  equation  by  the  common  flux  0  eq. 
(17)  is  obtained.  When  permeances  are  in  parallel,  the  total  flux 
is  the  sum  of  the  component  fluxes,  or 


eq 


(18a) 


Dividing  both  sides  of  this  equation  by  the  common  M,  eq.  (18)  is 
obtained. 


CHAP.  I]  FLUX  AND  MAGNETOMOTIVE  FORCE  17 

One  of  the  reasons  for  which  calculations  are  as  a  rule  more 
involved  and  less  accurate  in  the  magnetic  than  in  the  electric  cir- 
cuit is  that  there  is  no  magnetic  insulation  known,  and  therefore  the 
paths  of  the  flux  in  a  great  majority  of  cases  cannot  be  shaped  and 
confined  at  will.  The  student  will  appreciate,  therefore,  the  reason 
for  selecting  a  toroidal  ring  as  the  simplest  magnetic  circuit.  If 
the  winding  is  distributed  uniformly  there  is  no  tendency  for  mag- 
netic leakage,  except  for  a  very  small  amount  in  and  around  each 
wire.  With  almost  any  other  arrangement  of  a  magnetic  circuit 
there  is  a  difference  of  magnetic  potential,  or  an  m.m.f.  between 
various  parts  of  the  circuit,  and  part  of  the  flux  passes  directly 
through  the  path  of  the  least  resistance,  in  parallel  with  the  useful 
path.  A  familiar  example  of  this  is  the  magnetic  leakage  between 
the  adjacent  pole-tips  oian  electrical  machine  (Fig.29),  or  between 
the  coils  of  a  transformer  (Fig.  50) . 

The  conditions  in  a  magnetic  circuit  are  similar  to  those 
in  an  imperfectly  insulated  electric  circuit,  when  it,  together  with 
its  sources  of  e.m.f.,  is  immersed  in  a  conducting  liquid.  Part  of 
the  current  finds  its  path  through  the  liquid  instead  of  through 
the  conductors;  the  current  is  different  in  different  parts  of  the 
circuit,  and  the  calculations  are  much  more  involved  and  less 
accurate,  because  the  paths  of  the  current  in  an  unlimited  medium 
can  be  estimated  only  approximately. 

In  order  to  prevent  or  to  minimize  leakage  the  exciting  ampere- 
turns  should  be  distributed  over  the  whole  magnetic  circuit,  to 
each  part  in  proportion  to  its  reluctance.  Then  the  m.m.f.  is  con- 
sumed where  it  is  applied,  and  no  free  m.m.f.  is  left  for  leakage. 
Unfortunately,  such  an  arrangement  is  impracticable  in  most 
cases,  though  it  ought  to  be  approached  as  nearly  as  possible  (see 
Prob.  17  below). 

If  there  were  a  magnetic  insulator,  that  is,  a  substance  or  a 
combination  the  permeability  of  which  was  many  times  lower  than 
that  of  the  air,  it  would  be  a  great  boon  to  the  electrical  industry. 
It  would  then  be  possible  to  avoid  magnetic  leakage  by  insula- 
ting magnetic  circuits  as  perfectly  as  electric  circuits  are  insulated. 
The  absence  of  leakage  would  allow  a  reduction  in  the  size  of  the 
field  frames  and  exciting  coils  of  direct-  and  alternating-current 
machines.  It  would  also  permit  us  to  improve  the  voltage  regula- 
tion of  generators  and  transformers,  to  raise  the  power  factor  of 
induction  motors,  and  to  increase  considerably  their  overload 


18  THE  MAGNETIC  CIRCUIT  [ART.  9 

capacity;  it  would  also  largely  eliminate  sparking  in  commutating 
machines. 

Prob.  17.  A  long  iron  rod,  having  a  cross-section  of  9.3  sq.cm.,  is 
bent  into  a  circular  ring  so  that  the  ends  almost  touch  each  other.  The 
ring  is  wound  with  500  turns  of  wire,  the  winding  being  concentrated 
around  the  gap  to  minimize  the  leakage.  When  a  current  of  2.5  amperes 
is  sent  through  the  winding  a  flux  of  74.9  kilo-maxwells  is  established 
in  the  circuit.  Assuming  the  reluctance  of  the  iron  to  be  negligible, 
calculate  the  clearance  between  the  ends  of  the  rod. 

Ans.     Between  1.9  and  2.0  mm. 

Prob.  18.  What  is  the  length  of  the  air-gap  in  the  preceding  problem 
if  the  estimated  reluctance  of  the  iron  part  of  the  circuit  is  2  milli-rels? 

Ans.     1.7  mm. 

Prob.  19.  A  magnetic  circuit  consists  of  three  parts,  the  reluctances 
of  which  are  (R^  0.004  rel,  (R2  =  0.005  rel,  and  (R3  =  0.013  rel.  The 
paths  (R2  and  (R3  are  in  parallel  with  each  other  and  are  in  series  with 
(Ha.  What  is  the  total  permeance  of  the  circuit?  Ans.  131.4  perms. 

Prob.  20.  In  the  preceding  problem  let  (Hi  be  the  reluctance  of  the 
steel  frame  of  an  electric  machine,  (R2  be  that  of  two  air-gaps,  and  the 
armature,  and  (R3  the  leakage  reluctance  between  two  poles.  The  ratio 
of  the  total  flux  in  the  frame  to  the  useful  flux  through  the  armature 
is  called  the  leakage  factor  of  the  machine.  What  is  its  value  in  this 
case?  Ans.  1.38. 

Prob.  21.  Referring  to  the  two  preceding  problems  let  the  air-gap  be 
reduced  so  as  to  reduce  the  leakage  factor  to  1.2.  How  many  ampere- 
turns  will  be  required  to  produce  a  useful  flux  of  2.1  megalines  in  the 
magnetic  circuit  under  consideration?  Ans.  12,950. 

Prob.  22.  An  iron  ring  having  a  cross-section"  of  4  by  5  cms.  is  placed 
inside  of  a  hollow  ring.  This  ring  has  a  mean  diameter  of  32  cm.,  an 
axial  width  of  11  cm.,  and  a  radical  thickness  of  8  cm.  How  many 
ampere-turns  are  required  to  produce  a  total  flux  of  47  kilolines  (count- 
ing that  in  the  air  as  well  as  that  in  the  iron),  if  the  estimated  relative 
permeability  of  the  iron  is  1400?  Hint:  Let  the  average  flux  density  in 
the  air  be  Ba,  and  that  in  the  iron  be  B{.  We  have  two  simultaneous 
equations:  20£;  +  (88-20)#o=47,  and  £;/£a  =  1400.  Ans.  134. 

Prob.  23.  What  per  cent  of  the  total  flux  in  the  preceding  problem 
is  in  the  air?  Ans.  0.24  per  cent. 

Prob.  24.  Show  that  in  a  ring,  such  as  is  shown  in  Fig.  1,  the  flux 
density,  strictly  speaking,  is  not  uniform,  but  varies  inversely  as  the 
distance  from  the  center.  Solution:  Take  an  elementary  tube  of  flux 
of  a  radius  x.  The  magnetic  intensity  at  any  point  within  the  tube  is 
H=M/2nx,  and  the  flux  density,  according  to  eq.  (16),  B  =  /i.M/2nx. 

Prob.  25.  What  is  the  true  permeance  of  a  circular  ring  of  rectangular 
cross-section,  the  outside  diameter  of  which  is  Dlt  the  inside  diameter  D2, 
and  the  axial  width  ht  Solution:  The  permeance  of  an  infinitesimal 
tube  of  radius  x  is  cKP  =/j.hdx/2xx.  The  permeances  of  all  the  tubes 


CHAP.  I]  FLUX  AND  MAGNETOMOTIVE  FORCE  19 

are  in  parallel  and  should  be  added;  hence,  integrating  the  foregoing 
expression  between  the  limits  %Dl  and  %D2  we  get  :  (P  =  (/j.h/2rc)'Ln(Dl/D2)  . 
Prob.  26.  Show  that,  when  the  radial  thickness  b  of  a  ring  is  small  as 
compared  to  its  mean  diameter  D,  the  exact  expression  for  permeance, 
obtained  in  the  preceding  problem,  differs  but  little  from  the  approxi- 
mate value,  fthb/xD,  used  before.  Solution:  Using  the  expansion, 

..    and  putting 


we  get  ($>  =  (iJLhb/7:D)[l+%(b/Dy+%(b/D)4  +  .  .  .  ].  When  the  ratio  of 
b  to  D  is  small,  all  the  terms  within  the  brackets  except  the  first  one, 
can  be  neglected. 

Prob.  27.  Show  that  the  answer  to  prob.  11  is  2.1  per  cent  high  on 
account  of  the  density  being  assumed  there  as  uniform  throughout  the 
cross-section  of  the  ring. 


CHAPTER  II 
THE    MAGNETIC    CIRCUIT   WITH    IRON 

10.  The  Difference  between  Iron  and  Non-Magnetic  Materials. 

Steel  and  iron  differ  in  their  magnetic  properties  from  most  other 
known  materials  in  the  following  respects : 

(1)  The  permeability  of  steel  and  iron  is  several  hundred  and 
even  thousand  times  greater  than  that  of  non-magnetic  materials. 

(2)  The  permeability  of  steel  and  iron  is  not  constant,  but 
decreases  as  the  flux  density  increases. 

(3)  Changes    in   the    magnetization    of    steel    and  iron   are 
accompanied  by  some  sort  of  molecular  friction  (hysteresis)  with 
the  result  that  the  same  magnetomotive  force  produces  a  different 
flux  when  the  exciting  current  is  increasing  than  when  it  is  de- 
creasing (Fig.  7). 

Besides  iron,  the  four  adjacent  elements  in  the  periodic  system, 
viz.,  cobalt,  nickel,  manganese,  and  chromium,  are  slightly  mag- 
netic. Some  alloys  and  oxides  of  these  metals  show  considerable 
magnetic  properties.  Heusler  succeeded  in  producing  alloys  of 
manganese,  aluminum,  and  copper  which  are  strongly  magnetic. 
These  alloys  have  not  been  used  in  practice  so  far.1 

11.  Magnetization  Curves.  The  magnetic  properties  of  the  steel 
and  iron  used  in  the  construction  of  electrical  machinery  are  shown 
in  Figs.  2  and  3.     These  curves  are  called  magnetization  curves,  or 
B — H  curves]    sometimes    also    the    saturation    curves    of  iron. 
The  flux  density,  in  kilolines  per  square  centimeter  of  cross-sec- 
tion, is  plotted,  in  these  curves,  against  the  ampere-turns  per 
centimeter  length  of  the  magnetic  circuit  as  abscissae. 

The  student  may  conveniently  think  of  these  curves  as  represent- 

1  For  the  preparation  and  properties  of  Heusler's  alloys  see  Guthe  and 
Austin,  Bulletins  of  Bureau  of  Standards,  Vol.  2  (1906),  No.  2,  p.  297;  Dr.  C. 
P.  Steinmetz,  Electrical  World,  Vol.  55  (1910),  p.  1209;  Knowlton,  Physical 
Review,  Vol.  32  (1911),  p.  54. 

20 


CHAP.  II] 


MAGNETIC  CIRCUIT  WITH  IRON 


21 


ing  the  results  of  tests  on  samples  of  iron  in  ring  form,  as  in  Fig.  I.1 
The  current  in  the  exciting  coil  is  adjusted  to  a  certain  value,  and 
the  corresponding  value  of  the  flux  in  the  iron  ring  is  determined 
by  any  of  the  known  means,  for  instance,  by  a  discharge  through 


50  100  150    Scale  "B"     200 

10  20  30     Scale  "A"      40 

H  =  AMPERE-TURNS  PER  CENTIMETER 

Fig.  2 — Magnetization  in  steel  and  iron — castings  and  forgings. 

1  For  an  experimental  study  of  the  magnetic  circuit  with  iron  and  for 
practical  testing  of  the  magnetic  properties  of  steel  and  iron  see  Vol.  1, 
Chapters  6  and  7,  of  the  author's  Experimental  Electrical  Engineering. 


22  THE  MAGNETIC  CIRCUIT  [ART.  11 

a  secondary  coil  connected  to  a  calibrated  ballistic  galvanometer. 
The  exciting  ampere-turns  divided  by  the  average  length  of  the 
path  give  the  magnetic  intensity  H.  The  total  flux  divided  by  the 
cross-section  of  the  iron  path  gives  the  value  of  the  flux  density  B, 
which  is  plotted  as  an  ordinate  against  H  for  an  absissa.  Similar 
tests  are  made  for  other  values  of  H  and  B\  the  results  give  the 
magnetization  curve  of  the  material.  In  other  words,  a  magneti- 
zation curve  gives  the  relation  between  the  magnetomotive  force 
and  the  flux  for  a  unit  cube  of  the  material.  By  combining  unit 
cubes  in  series  and  in  parallel  a  relationship  is  established 
between  flux  and  ampere-turns  for  a  circuit  of  any  dimensions, 
made  of  the  same  material. 

The  curves  shown  in  Fig.  2  refer  to  the  following  materials:  (a) 
Cast  iron,  which  is  used  as  the  magnetic  material  in  the  stationary 
field  frames  of  direct-current  machines,  and  in  the  revolving-field 
spiders  of  low  speed  alternators.  It  is  evident  from  the  curves 
that  cast  iron  is  magnetically  much  inferior  to  steel ;  but  it  is  used 
on  account  of  its  lower  cost  and  ease  of  machining.  (6)  Cast  steel, 
which  is  used  for  pole  pieces,  plungers  of  electromagnets,  etc. 
It  is  used  also  for  the  field  frames  of  such  machines  in  which 
economy  of  weight  or  space  is  desired,  for  instance,  in  railway 
and  crane  motors,  and  in  machines  built  for  export,  (c)  Forged 
steel,  which  is  used  for  the  revolving  fields  of  turbo-alternators, 
on  account  of  the  considerable  mechanical  stresses  developed  in 
such  high  speed  machines  by  the  centrifugal  force. 

The  curves  in  Fig.  3  refer  to  carbon-steel  laminations  and  to 
silicon-steel  laminations.  The  former  is  used  in  the  armatures  of 
direct  and  alternating-current  machines,  the  latter  mainly  in  trans- 
formers. There  is  not  much  difference  between  the  two  kinds  with 
regards  to  their  B — H  curves,  but  silicon  steel  shows  a  much  lower 
loss  of  energy  due  to  hysteresis  and  eddy  currents  (see  Art.  20 
below).  A  material  of  much  higher  permeability  is  used  for 
armature  cores,  when  it  is  desired  to  use  very  high  flux  densities 
in  the  teeth.  A  magnetization  curve  for  such  steel  laminations  is 
shown  in  Fig.  28. 

For  convenience  and  accuracy  the  lower  part  of  each  curve  in 
Fig.  2  is  plotted  separately  to  a  larger  scale,  "  A,"  while  the  upper 
parts  are  plotted  to  a  smaller  scale,  "  B."  Thus,  Fig.  2  contains 
only  three  complete  magnetization  curves.  The  curve  for  silicon- 
steel  laminations  in  Fig.  3  is  also  plotted  to  two  different  scales, 


CHAP.  II]  MAGNETIC  CIRCUIT  WITH  IRON  23 

1000  2000  3000   Scale"C"  4000 


CQ 


1000  2000  3000    Scale  "C"  4000 

100  200  300    Scale  "B"  400 

10  20  30     Scale  "A"  40 

H  =  AMPERE-TURNS  PER  CENTIMETER 

FIG.  3. — MagnetizaticJb.  in  steel  laminations. 


24  THE  MAGNETIC  CIRCUIT  [ART.  12 

while  three  different  scales  are  used  for  the  carbon-steel  curve. 
The  values  of  H  at  very  low  flux  densities  are  unreliable  because 
in  reality  each  curve  has  a  point  of  inflexion  near  the  origin,  not 
shown  in  Figs.  2  and  3  (see  Fig.  7). 

The  curves  given  in  Figs.  2  and  3  represent  the  averages  of  many 
curves  obtained  from  various  sources.  The  iron  used  in  an  indi- 
vidual case  may  differ  considerably  in  its  magnetic  quality  from  the 
average  curve.  The  value  of  B  obtainable  with  a  given  H  depends 
to  a  large  degree  upon  the  chemical  constitution  of  the  specimen, 
impurities,  heat  treatment,  etc.  As  a  rule,  the  soft  and  pure  grades 
of  steel  are  magnetically  better,  that  is  to  say,  they  give  a  higher 
flux  density  for  the  same  magnetizing  force,  or,  what  is  the  same, 
they  possess  a  higher  permeability.  Annealing  improves  the 
magnetic  quality  of  iron,  while  punching,  hammering,  etc.,  lowers 
it.  Therefore,  the  laminations  used  in  the  construction  of  elec- 
trical machinery  are  usually  annealed  after  being  punched  into 
their  final  shape.  This  annealing  also  reduces  hysteresis  loss. 

12.  Permeability  and  Saturation.  Permeability  is  defined  in 
Chapter  I  as  the  permeance  of  a  unit  cube,  or,  according  to  eq. 
(16),  as  the  ratio  of  B  to  H.  The  two  definitions  are,  of  course, 
identical.  Therefore,  the  values  of  permeability  for  various  values 
of  B  are  easily  obtained  from  the  magnetization  curves.  For 
instance,  for  cast  steel,  at  B=  15  kilolines  per  sq.  cm.  the  magnetic 
intensity  H  is  26  ampere-turns  per  cm.,  so  that  JJL=  15000/26  =  577 
perms  per  cm.  cube.  This  is  the  value  of  the  absolute  permeabil- 
ity in  the  ampere-ohm-maxwell  system.  In  most  books  the  relative 
permeability  of  iron  is  employed,  referring  to  that  of  the  air  as 
unity.  Since  in  the  above-mentioned  system  //=1.25  for  air, 
the  relative  permeability  of  cast  steel  at  the  selected  flux  density 
is  577/1.25  =  461. 

In  practice,  the  calculations  of  magnetic  circuits  with  iron  are 
arranged  so  as  to  avoid  the  use  of  permeability  /*  altogether,  using 
the  B  —  H  curves  directly.  In  some  special  investigations,  how- 
ever, it  is  convenient  to  use  the  values  of  permeability,  and  also 
an  empirical  equation  between  PL  and  B.  For  instance,  see  the 
Standard  Handbook  for  Electrical  Engineers ;  the  topic  is  indexed 
"  permeability — curves/'  and  "  permeability — equation/'  These 
fji—B  curves  show  that  there  must  be  a  point  of  inflection  in 
the  B—H  curves  at  low  densities,  because  the  values  of  /*  reach 
their  maximum  at  a  certain  definite  density  instead  of  being  con- 


CHAP.  II]  MAGNETIC  CIRCUIT  WITH  IRON  25 

slant  for  the  lower  part  of  the  curves.  Such  would  be  the  case  if 
the  lower  parts  of  the  B—H  curves  were  straight  lines,  as  shown 
in  Figs.  2  and  3,  because  then  the  ratio  of  B  to  H  would  be  con- 
stant. However,  in  ordinary  engineering  work  the  lower  parts  of 
magnetization  curves  are  usually  assumed  to  be  straight  lines,  and 
the  permeability  constant. 

Three  parts  can  be  distinguished  in  a  B — H  or  magnetization 
curve:  the  lower  straight  part,  the  middle  part  called  the  knee  of 
the  curve,  and  the  upper  part,  which  is  nearly  a  straight  line.  As 
the  magnetic  intensity  H  increases,  the  corresponding  flux  density 
B  increases  more  and  more  slowly,  and  the  iron  is  said  to  approach 
saturation.  With  very  high  values  of  the  magnetic  intensity  H, 
say  several  thousand  ampere-turns  per  centimeter,  the  iron  is  com- 
pletely saturated  and  the  rate  of  increase  of  flux  density  with  H  is 
the  same  as  in  air  or  in  any  other  non-magnetic  material.  That  is 
to  say,  the  flux  density  B  increases  at  a  rate  of  1.257  kilolines  for 
each  kilo-ampere-turn  increase  in  H.  Such  is  the  slope  of  the  upper 
curve  in  Fig.  3. 

In  view  of  this  phenomenon  of  saturation  the  total  flux  density 
in  iron  can  be  considered  as  consisting  of  two  parts,  one  due  to  the 
presence  of  iron,  the  other  independent  of  it,  as  if  the  paths  of  the 
lines  of  force  were  in  air.  These  two  parts  are  shown  separately 
in  Fig.  4.  The  part  OA,  due  to  the  iron,  approaches  a  limiting 
value  B8)  where  the  iron  is  saturated.  The  part  OC,  not  due  to  the 
iron,  increases  indefinitely  in  accordance  with  the  straight  line 
law,  B  =  fj.H,  where  /*=  1.257.  The  curve  OD  of  total  flux  density 
resembles  in  shape  that  of  OA,  but  approaches  asymptotically 
a  straight  line  KL  parallel  to  OC. 

While  it  is  customary  to  speak  of  the  saturation  in  iron  as  being 
low,  high,  or  medium,  the  author  is  not  aware  of  any  generally 
recognized  method  of  expressing  the  degree  of  saturation  numeri- 
cally. It  seems  reasonable  to  define  per  cent  saturation  in  iron  with 
respect  to  the  flux  density  B8}  so  that,  for  instance,  the  per  cent 
saturation  at  the  point  N  is  equal  to  the  ratio  of  PN'  to  Bs.  This 
method  of  defining  saturation,  while  correct  theoretically,  pre- 
supposes that  the  ordinate  B8  is  known,  which  is  not  always  the 
case. 

The  percentage  saturation  of  a  machine  is  defined  in  Art.  58  of 
the  Standardization  Rules  of  the  American  Institute  of  Electrical 
Engineers  (edition  of  1910)  as  the  percentage  ratio  of  OQ  to  PN, 


26 


THE  MAGNETIC  CIRCUIT 


[AKT.  13 


QN  being  a  tangent  to  the  saturation  curve  at  the  point  N  under 
consideration.  An  objection  to  this  definition  is  that  according  to  it 
the  per  cent  saturation  does  not  approach  100  as  N  increases 
indefinitely;  on  the  contrary,  the  per  cent  saturation  gradually 
decreases  to  zero  beyond  a  certain  value  of  N.  This  is,  of  course, 
absurd.  Moreover,  the  foregoing  definition  of  the  Institute  refers 
explicitly  to  the  "  percentage  of  saturation  of  a  machine,"  and  it 
is  not  clear  whether  magnetization  curves  of  the  separate  materials 
are  included  in  it  or  not.  The  practical  advantage  of  this  definition 
as  compared  to  that  given  above  is  that  it  is  not  necessary  to 
know  the  value  of  B8. 


FIG.  4. — A  magnetization  curve  analyzed. 

13.  Problems  Involving   the  Use   of  Magnetization  Curves. 

The  following  problems  have  been  devised  to  give  the  reader  a  clear 
understanding  of  the  meaning  of  magnetization  curves,  and  to 
develop  fluency  in  their  use.  These  problems  lead  up  to  the 
magnetic  circuit  of  electric  machines  treated  in  Chapters  V  and 
VI.  With  almost  any  arrangement  of  a  magnetic  circuit  there 
is  some  leakage  or  spreading  of  the  lines  of  force,  which  is  difficult 
to  take  into  account  theoretically.  This  leakage  is  neglected  in 
most  of  the  problems  that  follow,  so  that  the  results  are  only 
approximately  correct.  Leakage  is  considered  more  in  detail  in 
Art.  40  below,  though  practical  designers  are  usually  satisfied  with 


CHAP.  II]  MAGNETIC  CIRCUIT  WITH  IRON  27 

estimating  it  from  the  results  of  previous  tests,  rather  than  to 
calculate  it  theoretically. 

Prob.  1.  Samples  of  cast  steel  are  to  be  tested  for  their  magnetic 
quality  up  to  a  density  of  19  kilolines  per  square  centimeter.  They  are 
to  be  in  the  form  of  rings,  20  cm.  average  diameter,  and  0.75  sq.cm. 
cross-section.  For  how  many  ampere-turns  should  the  exciting  winding 
be  designed,  and  what  is  the  lowest  permeance  of  the  circuit,  if  some 
specimens  are  expected  to  have  a  permeability  10  per  cent  lower  than 
that  according  to  the  curve  in  Fig.  2  ? 

Ans.     10.4  kiloampere-turns;  1.37  perm. 

Prob.  2.  Explain  the  reason  for  which  it  is  not  necessary  to  know  the 
cross-section  of  the  specimens  in  order  to  calculate  the  necessary  ampere- 
turns  in  the  preceding  problem. 

Prob.  3.  Some  silicon  steel  laminations  are  to  be  tested  in  the  form  of 
a  rectangular  bunch  20  by  2  by  1  cm.,  in  an  apparatus  called  a  permeam- 
eter.  The  net  cross-section  of  the  iron  is  90  per  cent  of  that  of  the 
packet.  It  is  found  for  a  sample  that  336  ampere-turns  are  required  to 
produce  a  flux  of  25.2  kilo-maxwells,  the  ampere-turns  for  the  air-gaps  and 
for  the  connecting  yoke  of  the  apparatus  being  eliminated.  How  does 
the  quality  of  the  specimen  compare  with  the  curve  in  Fig.  3? 

Ans.  The  permeability  of  the  sample  at  B  =  14  is  about  5  per  cent 
lower  than  that  according  to  the  curve. 

Prob.  4.  What  are  the  values  of  the  absolute  and  the  relative  per- 
meability and  reluctivity  of  the  sample  in  the  preceding  problem? 

Ans.  n  v 

relative   663  (numeric)  0.00151  (numeric) 

absolute  833  perms  per  cm.  cube     0.00120  rels  per  cm.  cube. 

Prob.  5.  What  is  the  maximum  permeability  of  cast  iron  according  to 
the  curve  in  Fig.  2?  Ans.  About  600  perms  per  cm.  cube. 

Prob.  6.  Mark  in  Figs.  2  and  3  vertical  scales  of  absolute  and  relative 
permeability,  so  that  values  of  permeability  could  be  read  off  directly  by 
laying  a  straight  edge  between  the  origin  and  the  desired  point  of  the 
magnetization  curve. 

Prob.  7.  What  is  the  percentage  of  saturation  in  carbon  steel  lamina- 
tions at  a  flux  density  of  20  kilo-maxwells  per  square  centimeter, 
according  to  both  definitions  given  in  Art.  12?  Ans.  92.5;  88.5. 

Prob.  8.  An  electromagnet  has  the  dimensions  (in  cm.)     shown   in 
Fig.  5;   the  core  is  made  of  carbon  steel  laminations  4  mm.  thick,  the 
lower  yoke  is  of  cast  iron.    The  length  of  each  air-gap  is  2  mm.;  each 
exciting  coil  has  450  turns.     What  is  the  exciting  current  for  a  useful 
flux  of  2.2  megalines  in  the  lower  yoke?    Neglect  the  magnetic  leakage 
between  the  limbs  of  the  electromagnet  (this  leakage  is  taken  into  consid- 
eration in  the  next  problem).    £eKMe^:(With  laminations  4  mm.  thick 
the  space  occupied  by  insulation  between  stampings  is  altogether  negligi-    i/ 
ble ;  therefore  the  flux  density  in  the  steel  is  the  same  as  in  the  air-gap,) 
and  is  equal  to  17.2  kl/sq.  cm.;  in  the  cast  iron  the  flux  density  is  11.5 


' 


28 


THE  MAGNETIC  CIRCUIT 


[ART.  13 


kl/sq.  cm.  One-half  of  the  average  length  of  the  path  in  the  steel  is 
37.3  cm.,  and  in  the  cast  iron  20.5  cm.  Hence,  with  reference  to  the 
magnetization  curves,  we  find  for  one-half  of  the  magnetic  circuit  (the 
other  half  being  identical,  it  is  sufficient  to  calculate  for  one-half) : 

amp.-turns  for  steel  core  65  X  37.3  =  2425 

amp  .-turns  for  one  air-gap  0.2  X  0.8  X 17200  =  2752 

amp.-turns  for  the  cast-iron  yoke  180  X  20.5  =  3690 


Total 


8867. 


Ans.    The  exciting  current  is  8867/450  =  19.7  amperes. 
r      Prob.  9.  In  the  solution  of  the  preceding  problem  the  effect  of  leakage 
is  disregarded.     It  is  found  by  experiments  on    similar  electromagnets 


FIG.  5. — An  electromagnet  (dimensions  in  centimeters). 


that  the  leakage  factor  is  equal  to  about  1.2,  that  is  to  say,  the  flux  in  the 
upper  yoke  is  20  per  cent  higher  than  that  in  the  lower  one.  This  means 
that  out  of  every  1200  fines  of  force  in  the  upper  yoke  1000  pass  through 
the  lower  yoke  as  a  part  of  the  useful  flux,  and  200  find  their  path  as  a 
leakage  through  the  air  between  the  limbs,  as  shown  by  the  dotted  lines. 
Calculate  the  exciting  current  required  in  the  preceding  problem,  assuming 
(a)  that  the  total  leakage  flux  is  concentrated  between  the  two  air-gaps 
along  the  line  aa;  (6)  that  it  is  concentrated  along  the  line  bb,  at  one- 
third  of  the  distance  from  the  bottom  of  the  exciting  coil,  that  is  6.33 
cm.  from  the  air-gaps. 

Ans.     (a)  44.2  amperes;  (6)  40  amperes. 

Prob.  10.  Show  that  it  is  more  correct  in  the  preceding  problem  to 
assume  the  leakage  flux  concentrated  at  one-third  of  the  distance  from 
the  bottom  of  the  exciting  coils,  than  at  the  center  of  the  coils. 


CHAP.  II]  MAGNETIC  CIRCUIT  WITH  IRON  29 

Prob.  11.  A  ring  of  forged  steel  has  such  dimensions  that  the  average 
length  of  the  lines  of  force  is  70  cm.  The  ring  has  an  air-gap  of  1.5  mm.,  and 
is  provided  with  an  exciting  winding  concentrated  near  the  air-gap  so  as 
to  minimize  the  leakage.  What  is  the  flux  density  at  an  m.m.f.  of  4000 
ampere- turns?  First  Solution :  Assume  various  values  of  B,  calculate  the 
corresponding  values  of  the  ampere-turns,  until  the  value  of  B  is  found,  for 
which  the  required  excitation  is  4000  ampere-turns  (solution  by  trials). 
Second  solution:  Let  the  unknown  density  be  B  and  the  corresponding 
magnetic  intensity  in  the  steel  be  H.  The  required  excitation  for  the  steel 
is  then  7QH,  and  for  the  air-gap  0.15X0.8X10005  =  1205  ampere-turns. 
Therefore, 

70#  + 1205 =4000. 

The  values  of  B  and  H  must  satisfy  this  equation  of  a  straight  line,and 
besides  they  must  be  related  to  each  other  by  the  magnetization  curve 
for  steel  forgings  (Fig.  2).  Hence,  B  and  H  are  determined  by  the  intersec- 
tion of  the  straight  line  and  the  curve.  The  straight  line  is  determined  by 
two  of  its  points ;  for  instance,  when  H = 40,  B  =  10 ;  when  H  =  24,  B  =  19.3. 
Drawing  this  line  in  Fig.  2  we  find  that  the  point  of  intersection  corre- 
sponds to  B  =  16.3. :  Ans.  16.3  kilolines  per  sq.  cm. 

Prob.  12.  Solve  the  preceding  problem,  assuming  the  ring  to  be  made 
of  silicon  steel  laminations:  10  per  cent  of  the  space  is  taken  by  the 
insulation  between  the  laminations. 

Ans.    Flux  density  in  the  laminations  is  15.2  kl/sq.  cm. 

Prob.  13.  In  a  complex  magnetic  circuit,  an  air-gap  3  mm.  long  and 
26  sq.  cm.  in  cross-section  is  shunted  by  a  cast-iron  rod  14  cm.  long  and 
10  sq.  cm.  in  cross-section.  WTiat  is  the  number  of  ampere-turns  neces- 
sary for  producing  a  total  flux  of  215  kilolines  through  the  two  paths  in 
parallel,  and  what  is  the  reluctance  of  the  rod  per  centimeter  of  its  length 
under  these  conditions?  Ans.  1160  ampere-turns;  0.933  milli-rel. 

Prob.  14.  The  magnetic  flux  in  a  closed  iron  core  must  increase  and 
decrease  according  to  a  straight-line  law  with  the  time,  then  reverse  and 
increase  and  decrease  according  to  the  same  law  in  the  opposite  direction. 
Show  the  general  shape  of  the  curve  of  the  exciting  current,  neglecting 
the  effect  of  hyteresis. 

Prob.  15.  Show  that  if  in  the  preceding  problem  the  flux  varies  accord- 
ing to  the  sine  law  the  curve  of  the  exciting  current  is  a  peaked  wave. 
Show  how  to  determine  the  shape  of  this  curve  from  a  given  magnetiza- 
tion curve  of  the  material.  This  problem  has  an  application  in  the  calcu- 
lation of  the  exciting  current  in  a  transformer. 

Prob.  16.  In  the  magnetic  circuit  shown  in  Fig.  6  the  useful  flux 
passes  through  the  air-gap  between  the  two  steel  poles;  a  part  of  the  flux 

1  The  student  will  see  from  the  solution  of  this  problem  that  in  the  case 
of  a  series  magnetic  circuit  it  is  much  easier  to  find  the  m.m.f.  required  for 
a  given  flux  than  vice  versa.  On  the  other  hand,  in  the  case  of  two  mag- 
netic paths  in  parallel  (such  as  in  prob.  13),  it  is  easier  to  find  the  flux  for  a 
given  m.m.f. 


30 


THE   MAGNETIC  CIRCUIT 


[ART.  13 


is  shunted  through  the  cast-iron  part  of  the  circuit.  At  low  saturations  a 
considerable  part  of  the  total  flux  is  shunted  through  the  cast-iron  part, 
but  as  the  flux  density  increases  the  cast  iron  becomes  saturated,  and  a 
larger  and  larger  portion  of  the  flux  is  deflected  into  the  air-gap.  What 
percentages  of  the  total  flux  in  the  yoke  are  shunted  through  the  cast  iron 
when  the  flux  density  in  the  air-gap  is  1  kl/sq.  cm.  and  7  kl/sq.  cm. 
respectively?  Solution :  When  the  flux  density  in  the  air-gap  is  1  kilo- 
line  per  sq.  cm.  the  m.m.f.  across  the  gap  is  1000X0.8X0.5  =  400  ampere- 
turns.  The  flux  density  in  the  steel  poles  is  2  kl/sq.  cm.,  and  the  required 
m.m.f.  in  them  is  about  16  ampere-turns.  Therefore,  the  total  m.m.f. 
across  AC  and  consequently  across  the  cast-iron  part  is  416  ampere-turns, 


Sq.  Cm. 


35  Sq.  Cm. 
Cas.t  Sleel 


FIG.  6. — A  complex  magnetic  circuit. 

or  H  =  24.5  ampere-turns  per  centimeter  of  length  of  the  path  in  the  cast 
iron.  Thisvalue  of //corresponds  on  the  magnetization  curve  toB  =  6kl/sq. 
cm. ;  hence,  the  total  n\ux  in  the  cast  iron  is  72  kl.  The  flux  in  the  yoke  is 
60  +  72  =  132  kl.,  and  the  percentage  in  the  cast-iron  shunt  is  72/132  or 
about  55  per  cent.  Similarly,  it  is  found  that,  when  the  flux  density  in  the 
air-gap  is  7/kl  sq.  cm.,  about  25  per  cent  of  the  flux  is  shunted  through 
the  cast-iron  part.  The  foregoing  arrangement  illustrates  the  principle 
used  in  some  practical  cases,  when  it  is  desired  to  modify  the  relation 
between  the  flux  and  the  magnetomotive  force,  by  providing  a  highly 
saturated  magnetic  path  in  parallel  with  a  feebly  saturated  one. 

Ans.     55  per  cent  and  25  per  cent  approximately. 
Prob.  17.    Indicate  how  the  preceding  problem  can  be  solved  if  the 
cast-iron  part  were  provided  with  a  small  clearance  of  say  1  mm.     Hint: 
See  the  second  solution  to  problem  1 1 . 


CHAP.  II]  MAGNETIC  CIRCUIT  WITH  IRON  31 

Prob.  18.  What  is  the  length  of  the  yoke  in  Fig.  6  if  the  exciting  cur- 
rent increases  12  times  when  the  flux  density  in  the  air-gap  increases  from 
1  to  7  kl/sq.  cm.?  Hint:  If  Hx  and  H7  are  the  known  magnetic  intensi- 
ties in  the  yoke,  corresponding  to  the  two  given  densities,  and  x  is  the 
unknown  length  of  the  yoke,  we  have,  using  the  values  obtained  in  the 
solution  of  problem  15:  (H1x+416)12  =  H7x+3090. 

Ans.    About  1.2  m. 


CHAPTER  III 
HYSTERESIS  AND  EDDY  CURRENTS  IN  IRON 

14.  The  Hysteresis  Loop.  Steel  and  iron  possess  a  property 
of  retaining  part  of  their  magnetism  after  the  external  magnetomo- 
tive force  which  magnetized  them  has  been  removed.  Therefore, 
the  magnetization  or  the  B-H  curve  of  a  sample  depends  some- 
what upon  the  magnetic  state  of  the  specimen  before  the  test.  This 
property  of  iron  is  called  hysteresis.  The  curves  shown  in  Figs.  2 
and  3  refer  to  the  so-called  virgin  state  of  the  materials,  which  state 
is  obtained  by  thoroughly  demagnetizing  the  sample  before  the 
test.  A  piece  of  iron  can  be  reduced  to  the  virgin  state  by  placing 
it  within  a  coil  through  which  an  alternating  current  is  sent,  and 
gradually  reducing  the  current  to  zero.  Instead  of  changing  the 
current,  the  sample  can  be  removed  from  the  coil. 

Let  a  sample  of  steel  or  iron  to  be  tested  be  made  into  a  ring 
and  provided  with  an  exciting  winding,  as  in  Fig.  1.  Let  it  be 
thoroughly  demagnetized;  in  other  words,  let  its  residual  mag- 
netism be  removed ;  then  let  the  ring  be  magnetized  gradually  or  in 
steps  to  a  certain  value  of  the  flux  density.  Let  OA  in  Fig.  7  rep- 
resent the  virgin  magnetization  curve,  that  is  to  say  the  relation 
between  the  calculated  values  of  B  and  H  from  this  test,  and  let 
PA  be  the  highest  flux  density  obtained.  If  now  the  magnetizing 
current  be  gradually  reduced,  the  relation  between  B  and  H  is  no 
more  represented  by  the  curve  OA,  but  by  another  curve,  such  as 
AC;  this  is  because  of  the  above-mentioned  property  of  iron  to 
retain  part  of  its  magnetism.  When  the  current  is  reduced  to 
zero,  the  specimen  still  possesses  a  residual  flux  density  OC.  Let 
the  current  now  be  reversed  and  increased  in  the  opposite  direc- 
tion, until  H  reaches  the  negative  value  OF,  at  which  no  magnetic 
flux  is  left  in  the  sample.  The  value  of  H=OF  is  called  the  coer- 
cive force.  When  the  magnetic  intensity  reaches  the  negative  value 
of  OP'  =  OP,  experiment  shows  that  the  magnetic  density  P'A'  in 
the  sample  is  equal  and  opposite  to  PA. 

32 


CHAP.  Ill] 


HYSTERESIS  AND  EDDY  CURRENTS 


33 


Let  now  the  exciting  current  be  again  decreased,  reversed  and 
increased  to  its  former  maximum  value  corresponding  to  H=OP. 
It  will  be  found  that  the  relation  between  B  and  H  follows  a  differ- 
ent though  symmetrical  curve,  A'C'F'A,  which  connects  with  the 
upper  curve  at  the  point  A.  The  complete  closed  curve  is  called 
the  hysteresis  loop-,  a  sample  of  iron  which  has  been  subjected  to  a 
varying  magnetomotive  force  as  described  before,  is  said  to  have 
undergone  a  complete  cycle  of  magnetization.  If  the  same  cycle 


FIG.  7. — A  hysteresis  loop. 

be  repeated  any  number  of  times,  the  curve  between  B  and  H 
remains  the  same,  as  long  as  the  physical  properties  of  the  sample 
remain  unchanged. 

The  lower  half  of  the  hysteresis  loop  is  identical  with  the 
inverted  upper  half,  so  that  the  residual  flux  density  OC'  =  OC, 
and  the  coercive  force  OF'  =  OF.  The  shape  of  the  loop  for  a 
given  sample  is  completely  determined  by  the  maximum  ordinate 
AP,  or  the  maximum  excitation  OP.  If  the  excitation  be  carried 
further,  for  instance,  to  the  point  D  on  the"  virgin  curve,  the 
hysteresis  loop  would  be  larger,  beginning  at  the  point  D,  and 
would  be  similar  in  its  general  shape  to  the  loop  shown  in  Fig.  7. 


34  THE  MAGNETIC  CIRCUIT  [ART.  15 

A  piece  of  iron  can  also  be  carried  through  a  hysteresis  cycle 
mechanically.  Thus,  instead  of  changing  the  excitation,  the 
sample  may  be  moved  to  a  weak  field,  reversed,  and  returned  to 
its  original  location.  The  relation  between  B  and  H,  however, 
will  be  the  same  in  either  case. 

An  important  feature  of  the  hysteresis  cycle  is  that  it  requires 
a  certain  amount  of  energy  to  be  supplied  by  the  magnetizing 
current,  or  by  the  mechanism  which  reverses  the  iron  with 
respect  to  the  field.  It  is  proved  in  Art.  16  below  that  this  energy 
per  cubic  unit  of  iron  is  proportional  to  the  area  of  the  hysteresis 
loop.  This  energy  is  converted  into  heat  in  the  iron,  and  therefore 
from  the  point  of  view  of  the  electromagnetic  circuit  represents 
a  pure  loss.  If  the  cycles  of  magnetization  are  performed  in 
sufficiently  rapid  succession,  for  instance  by  using  alternating 
current  in  the  exciting  winding,  the  temperature  of  the  iron  rises 
appreciably. 

The  phenomenon  of  hysteresis  is  irreversible;  that  is  to  say, 
it  is  impossible  to  make  a  piece  of  iron  to  undergo  a  cycle  of  mag- 
netization in  the  direction  opposite  to  that  indicated  by  arrow- 
heads, in  Fig.  7.  If  it  were  reversible  the  loss  of  energy  occasioned 
by  performing  the  cycle  in  one  direction  could  be  regained  by 
performing  it  in  the  opposite  direction.  In  this  respect  the 
hysteresis  cycle  differs  materially  from  the  theoretical  reversible 
cycles  studied  in  thermodynamics,  and  reminds  one  of  an  irre- 
versible thermodynamic  cycle,  in  which  friction  or  sudden  expan- 
sion is  present. 

15.  An  Explanation  of  Saturation  and  Hysteresis  in  Iron. 
While  the  physical  nature  of  magnetism  is  at  present  unknown, 
there  is  sufficient  evidence  that  the  magnetization  of  iron  is 
accompanied  by  some  kind  of  molecular  change.  Let  us  assume, 
in  accordance  with  the  modern  electronic  theory,  that  there  is  an 
electric  current  circulating  within  each  molecule  of  iron,  due  to  the 
orbital  motion  of  one  or  more  electrons  within  the  molecule.  Each 
molecule  represents,  therefore,  a  minute  electromagnet  acted  upon 
by  other  molecular  electromagnets.  In  the  neutral  state  of  a  piece 
of  iron,  the  grouping  of  the  molecules  is  such  that  the  currents 
are  distributed  in  all  possible  planes,  and  the  external  magnetic 
action  is  zero.  Under  the  influence  of  an  external  magnetomotive 
force  the  molecules  are  oriented  in  the  same  way  that  small  mag- 
netic needles  are  deflected  by  an  external  magnetic  field.  With 


CHAP.  Ill]  HYSTERESIS  AND  EDDY  CURRENTS  35 

small  intensities  of  the  external  field,  the  molecules  of  iron  return 
into  their  original  stable  positions  as  soon  as  the  external  m.m.f. 
is  removed;  when,  however,  the  external  magnetic  intensity 
becomes  considerable  some  of  the  molecules  turn  violently  and 
assume  new  groupings  of  stable  equilibrium.  Therefore,  when 
the  external  m.m.f.  is  removed,  there  is  some  intrinsic  magneti- 
zation left,  and  we  have  the  phenomenon  of  residual  mag- 
netism. 

With  an  ever-increasing  external  m.m.f.,  more  and  more  of  the 
molecules  are  oriented  so  that  their  m.m.fs.  are  in  the  same  direc- 
tion as  the  external  field,  the  iron  then  approaching  saturation. 
Any  further  increase  in  the  flux  density  is  then  mainly  due  to  the 
flux  between  the  molecules,  the  same  as  in  any  non-magnetic 
medium. 

According  to  the  foregoing  theory,  an  external  m.m.f.  turns 
the  internal  m.m.fs.  into  more  or  less  the  same  direction;  these 
m.m.fs.  then  help  to  establish  the  flux  in  the  intermolecular  spaces 
which  are  much  greater  than  the  molecules  themselves.  There- 
fore, the  higher  flux  density  in  iron  is  not  due  to  a  greater  permea- 
bility of  the  iron  itself,  but  to  an  increased  m.m.f.  It  is  never- 
theless permissible,  for  practical  purposes,  to  speak  of  a  higher 
permeability  of  the  iron,  disregarding  the  internal  m.m.fs.,  and 
considering  the  permeability,  according  to  eq.  (16),  as  the  ratio 
of  the  flux  density  to  the  externally  applied  magnetic  intensity. 

The  foregoing  theory  explains .  also  the  general  character  of 
the  permeability  curve  of  iron.  With  very  small  values  of  H 
the  molecules  of  a  piece  of  iron  are  oriented  but  very  little,  but 
are  rapidly  oriented  more  and  more  as  H  is  increased.  There- 
fore, for  small  values  of  H,  /j.  must  be  expected  to  increase  with  H. 
On  the  other  hand,  when  the  saturation  is  very  high,  an  increase 
in  H  changes  B  but  little,  because  practically  all  of  the  available 
internal  m.m.fs.  have  been  utilized.  Therefore,  for  large  values 
of  H,  jj.  decreases  with  increasing  H.  Consequently,  there  is  a 
value  of  H  for  which  /*  is  a. maximum.  This  is  the  actual  shape  of 
permeability  curves  (see  for  instance  the  reference  to  the  Standard 
Handbook  given  in  Art.  12  above). 

The  phenomenon  of  magnetization  is  irreversible  because  the 
changes  from  one  stable  grouping  of  molecules  to  the  next  are 
sudden.  Each  molecule,  in  changing  to  a  new  grouping,  acquires 
kinetic  energy,  and  oscillates  about  its  new  position  of  equilib- 


36 


THE  MAGNETIC  CIRCUIT 


[ART.  15 


rium  until  the  energy  is  dissipated  by  being  converted  into  heat. 
This  heat  represents  the  loss  of  energy  due  to  hysteresis. 

This  theory  of  saturation  and  hysteresis  is  due  originally 
to  Weber,  and  has  been  improved  by  Ewing,  who  has  shown 
experimentally  the  possibility  of  various  stable  groupings  of 
a  large  number  of  small  magnets  in  a  magnetic  field.  By  varying 
the  applied  m.m.f.  he  obtained  a  curve  similar  to  the  hysteresis 
loop  of  a  sample  of  iron.  For  further  details  of  this  theory  see 
Ewing,  Magnetic  Induction  in  Iron  and  other  Metals  (1892), 
Chapter  XI. 

The  following  analogy  is  also  useful,  Let  a  body  Q  (Fig.  8), 
rest  on  a  support  and  be  held  in  its  central  position  by  two  springs 


FIG.  8. — A  mechanical  analogue  to  hysteresis. 

S,  S,  which  can  work  both  under  tension  and  under  compression. 
Let  this  body  be  made  to  move  periodically  to  the  right  and  to 
the  left  of  its  central  position,  under  the  influence  of  an  alterna- 
ting external  force  H.  Call  B  the  deflections  of  the  body  from 
its  middle  position.  The  relation  between  B  and  H  is  then  similar 
to  the  hysteresis  loop  in  Fig.  7,  provided  that  there  is  some 
friction  between  the  body  Q  and  its  support,  and  provided  that 
the  springs  offer  in  proportion  more  resistance  when  distorted 
greatly  than  when  distorted  slightly. 

Starting  with  the  neutral  position  of  the  body  let  a  gradually 
increasing  force  H  be  applied  which  moves  the  body  to  the  right. 
This  corresponds  to  the  virgin  curve  in  Fig.  7,  except  that  this 
simple  analogy  does  not  account  for  the  inflection  in  the  virgin 
curve  near  the  origin.  Let  then  the  force  H  be  gradually  reduced, 
allowing  the  springs  to  bring  Q  nearer  the  center.  When  the 


CHAP.  Ill]  HYSTERESIS  AND  EDDY  CURRENTS  37 

external  force  is  entirely  removed,  the  body  is  still  somewhat  to 
the  right  of  its  central  position,  because  the  friction  balances  part 
of  the  tension  of  the  springs.  Here  we  have  something  analogous 
to  residual  magnetism  and  to  the  part  AC  of  the  hysteresis  loop. 
A  finite  force  H  is  required  in  the  negative  direction  to  bring  Q 
to  the  center.  This  force  corresponds  to  the  coercive  force  of  a 
piece  of  iron. 

By  following  this  analogy  through  the  complete  cycle  one 
can  show  that  a  loop  is  obtained  similar  to  a  hysteresis  loop. 
Also,  it  can  be  shown  that  the  phenomenon  is  irreversible,  and 
that  total  work  done  by  the  force  H  is  equal  to  the  work  of  friction. 
Moreover  there  is  a  periodic  interchange  of  energy  between  the 
springs  and  the  source  of  the  force  H}  and  the  net  loss  of  energy 
is  represented  by  the  area  of  the  loop  corresponding  to  Fig.  7. 

Prob.  1.  An  iron  ring  is  thoroughly  demagnetized,  and  then  the  cur- 
rent in  the  exciting  winding  is  varied  in  the  following  manner:  It  is 
increased  gradually  from  zero  to  1  ampere  and  is  then  reduced  to  zero. 
After  this,  the  current  is  increased  to  2  amperes  in  the  same  direction,  and 
again  reduced  to  zero.  Then  the  current  is  increased  to  3  amperes  again  in 
the  same  direction,  and  reduced  to  zero,  etc.  Draw  roughly  the  general 
character  of  the  B-H  curve,  taking  the  hysteresis  into  consideration. 
Hint :  First  study  a  similar  process  on  the  mechanical  analogy  shown  in 
Fig.  8.1 

Prob.  2.  A  piece  of  iron  is  made  to  undergo  a  magnetization  process 
from  the  point  A  (Fig.  7)  to  a  point  between  F  and  Af  such  that,  when 
subsequently  the  exciting  circuit  is  opened,  the  ascending  branch  of  the 
hysteresis  curve  comes  to  the  origin.  Show  that  such  a  process  does  not 
bring  the  iron  into  the  neutral  virgin  state,  in  spite  of  the  fact  that  5  =  0 
for  H  =  Q.  Hint :  Consider  the  further  behavior  of  the  iron  for  positive 
and  negative  values  of  H . 

Prob.  3.  A  millivoltmeter  is  connected  to  the  high-tension  terminals 
of  a  transformer,  and  the  current  in  the  low-tension  winding  is  varied  in 
such  a  way  as  to  keep  the  voltage  constant :  Show  that  the  curve  of  the 
current  plotted  against  time  is  proportional  to  the  hysteresis  loop  of  the 
core.  Hint:  Since  d@/dt  is  constant,  0  is  proportional  to  the  time. 

Prob.  4.  The  magnetic  flux  density  in  an  iron  core  is  to  vary  with  the 
time  according  to  the  sine  law.  Plot  to  time  as  abscissae  the  instantane- 
ous values  of  the  exciting  ampere-turns  per  centimeter  length  of  the  core 
from  an  available  hysteresis  loop,  and  show  that  the  wave  of  the  exciting 
current  is  not  a  sine  wave  and  is  unsymmetrical.  Note:  This  problem 
has  an  application  in  the  calculation  of  the  exciting  current  of  a  trans- 
former; see  Art.  33  below. 

1  A  solution  of  this  and  of  the  next  problem  will  be  found  in  Chapter  V 
of  Ewing's  Magnetic  Induction  in  Iron  and  other  Metals,  1892. 


38  THE  MAGNETIC  CIRCUIT  [ART.  16 

16.  The  Loss  of  Energy  per  Cycle  of  Magnetization.  When  a 
magnetic  flux  is  maintained  constant  the  only  energy  supplied 
from  the  source  of  electric  power  is  that  converted  into  the  i2r 
heat  in  the  exciting  winding;  no  energy  is  necessary  to  maintain 
the  magnetic  flux.  This  is  an  experimental  fact,  fundamental  in 
the  theory  of  magnetic  phenomena.  When,  however,  the  flux 
is  made  to  vary,  by  varying  the  exciting  ampere-turns  or  the 
reluctance  of  the  magnetic  circuit,  electromotive  forces  are 
induced  in  the  magnetizing,  winding  by  the  changing  flux. 
A  transfer  of  energy  results  between  the  electric  and  the  magnetic 
circuits. 

Beginning,  for  instance,  at  the  point  A  of  the  cycle  (Fig.  7), 
and  going  toward  C,  the  flux  is  forced  to  decrease.  According 
to  Faraday's  law,  the  e.m.f .  induced  by  this  flux  in  the  magnetiz- 
ing winding  is  such  as  to  resist  the  change,  i.e.,  it  tends  to  main- 
tain the  current.  Therefore,  during  the  part  AC  of  the  cycle 
energy  is  supplied  from  the  magnetic  to  the  electric  circuit. 
This  shows  that  energy  is  stored  in  a  magnetic  field.  During  the 
part  CFA'  of  the  hysteresis  loop  energy  is  supplied  from  the 
electric  to  the  magnetic  circuit,  because  at  the  point  C,  the 
current  is  reversed  and  becomes  opposed  to  the  e.m.f.  The 
other  half  of  the  cycle  being  symmetrical,  with  the  flux  and  the 
current  reversed,  energy  is  returned  to  the  electric  circuit  during 
the  part  A'C'  of  the  cycle,  and  is  again  accumulated  in  the 
magnetic  circuit  during  the  part  C'F'A. 

If  the  part  AC  of  the  cycle  were  identical  with  C'F'A,  and  the 
part  A'C'  were  identical  with  CFA',  the  amounts  of  energy  trans- 
ferred both  ways  would  be  the  same,  and  there  would  be  no  net 
loss  of  energy  at  the  end  of  the  cycle.  In  reality  the  two  parts 
are  different;  the  amounts  of  energy  returned  from  the  magnetic 
circuit  to  the  electric  circuit  in  the  parts  AC  and  A'C'  are  smaller 
than  the  amounts  supplied  by  the  electric  circuit  in  the  parts 
CFA'  and  C'F'A.  This  is  because  the  last  two  parts  of  the  curve 
are  more  steep  than  the  first  two,  and  consequently  the  induced 
e.m.fs.  are  larger  for  the  same  values  of  the  current.  The  net 
result  is  therefore  an  input  of  energy  from  the  electric  into  the 
magnetic  circuit,  this  energy  being  converted  into  heat  in  the  iron. 
No  such  effect  is  observed  with  non-magnetic  materials,  because 
the  two  branches  of  a  complete  B-H  cycle  coincide  with  a  straight 
line  passing  through  the  origin. 


CHAP.  Ill]  HYSTERESIS  AND  EDDY   CURRENTS  39 

To  prove  that  the  energy  lost  per  cubic  unit  of  iron  per  cycle 
of  magnetization  is  represented  by  the  area  of  the  hysteresis  loop, 
we  first  write  down  the  expression  for  the  energy  returned  to  the 
electric  circuit  during  an  infinitesimal  change  of  flux  in  the 
part  AC  of  the  cycle.  Let  the  flux  in  the  ring  at  the  instant  under 
consideration  be  0  webers,  and  the  magnetomotive  force  ni  amp- 
ere-turns, where  i  is  the  instantaneous  value  of  the  current,  and  n 
is  the  total  number  of  turns  on  the  exciting  winding.  The  instan- 
taneous induced  e.m.f  .,  due  to  a  decrease  of  the  flux  by  d<P  during 
an  infinitesimal  element  of  time  dt  seconds,  is  e=  —ndtf>/dtvolt. 
The  sign  minus  is  necessary  because  e  is  positive  (in  the  direction 
of  the  current)  when  dd>  is  negative,  that  is  to  say,  when  the  flux 
decreases.  The  electric  energy  corresponding  to  this  voltage  is 

dW  =  eidt=  —nid$  watt-seconds  (joules). 

Hence,  the  total  energy  returned  to  the  electric  circuit  during  the 
part  AC  of  the  cycle  is- 

W=  -  C'nidQ, 

•/  A 

or,  interchanging  the  limits  of  integration, 

W=  C   nid$. 

^c 

Since  all  the  parts  of  the  ring  undergo  the  same  process,  and 
the  curve  in  Fig.  7  is  plotted  for  a  .unit  cube  of  the  material,  it  is 
of  interest  to  find  the  loss  of  energy  per  cubic  centimeter  of  mate- 
rial. If  &  is  the  cross-section  and  I  the  mean  length  of  the  lines 
of  force  in  the  iron,  we  have  that  the  volume 

V  =  Sl  cubic  centimeters. 

Dividing  the  expression  for  the  energy  by  this  equation,  we  find 
that  the  energy  hi  watt-seconds  per  cubic  centimeter  of  iron  is 


W/V-.f 


(19) 


where  H  is  in  ampere-turns  per  centimeter,  and  B  is  in  webers 
per  square  centimeter. 

But  HdB  is  the  area  of  an  infinitesimal  strip,  such  as  is  shown 
by  hatching  in  Fig.  7.     Consequently,  the  right-hand  side  of  eq. 


40  THE  MAGNETIC   CIRCUIT  [ART.  17 

(19)  represents  the  area  of  the  figure  ACQ,  which  is  therefore  a 
measure  for  the  energy  transferred  to  the  electric  circuit,  per  cubic 
centimeter.  In  exactly  the  same  way  it  can  be  shown  that  the 
energy  supplied  to  the  magnetic  circuit  during  the  part  C'A  of  the 
cycle  is  represented  by  the  area  AC'Q.  Hence  the  net  energy 
loss  for  the  part  of  the  cycle  to  the  right  of  the  axis  of  ordinates 
is  represented  by  the  area  ACC'A.  Repeating  the  same  reasoning 
for  the  left-hand  side  of  the  loop  it  will  be  seen  that  the  total 
energy  loss  per  cycle  of  magnetization  per  cubic  centimeter  of 
material  is  represented  by  the  area  AC  A' C'A  of  the  hysteresis 
loop.  For  a  given  material,  this  area,  and  consequently  the  loss, 
is  a  function  of  the  maximum  flux  density  PA,  and  increases  with 
it  according  to  a  rather  complicated  law.  Two  empirical  formulae 
for  the  loss  of  energy  as  a  function  of  the  density  are  given  in  Art. 
20  below. 

In  the  problems  that  follow  the  weight  of  one  cubic  decimeter 
of  solid  carbon  steel  is  taken  to  be  7.8  kg.,  and  that  of  the  alloyed 
or  silicon  steel  7.5  kg.  The  weight  of  one  cubic  decimeter  of 
assembled  carbon  steel  laminations  is  taken  as  0.9X7.8  =  7  kg., 
and  that  of  silicon  steel  laminations  as  0.9X7.5  =  about  6.8  kg. 

Prob.  5.  A  hysteresis  loop  is  plotted  to  the  following  scales :  abscissae 
1  cm.  =  10  amp  .-turns /cm. ;  ordinates,  1  cm.  =  l  kilo-maxwell/sq.  cm. ;  the 
area  of  the  loop  is  found  by  a  planimeter  to  be  72  sq.  cm.  What  is  the 
loss  per  cycle  per  cubic  decimeter  of  iron? 

Ans.  7.  2  watt-seconds  (joules). 

Prob.  6.  The  hysteresis  loop  mentioned  in  the  preceding  problem  was 
obtained  from  an  oscillographic  record  at  a  frequency  of  60  cy.,  with  a  sam- 
ple of  iron  which  weighed  9.2  kg.  What  was  the  power  lost  in  hysteresis 
in  the  whole  ring?  Ans.  510  watts. 

Prob.  7.  The  stationary  coil  of  a  ballistic  electro-dynamometer  is  con- 
nected in  series  with  the  exciting  electric  circuit  (Fig.  1) ;  the  moving  coil 
is  connected  through  a  high  resistance  to  a  secondary  winding  placed  on 
the  ring.  The  exciting  current  is  brought  to  a  certain  value,  and  then  the 
current  is  reversed  twice  in  rapid  succession,  in  order  that  the  iron  may 
undergo  a  complete  magnetization  cycle.  Show  that  the  deflection  of  the 
electro-dynamometer  is  a  measure  for  the  area  of  the  hysteresis  loop. 
Hint:  HdB=H(dB/dt)  ctt=Const.  X  iedt.1 

17.  Eddy  Currents  in  Iron.  Iron  is  an  electrical  conductor; 
therefore  when  a  magnetic  flux  varies  in  it,  electric  currents  are 

1  Searle,  "The  Ballistic  Measurement  of  Hysteresis,"  Electrician,  Vol. 
49,  1902,  p.  100. 


CHAP.  Ill]  HYSTERESIS  AND  EDDY  CURRENTS  41 

induced  along  closed  paths  of  least  resistance  linked  with  the  flux. 
These  currents  permeate  the  whole  bulk  of  the  iron  and  are  called 
eddy  or  Foucault  currents.  Eddy  currents  cause  a  loss  of  energy 
which  must  be  supplied  either  electrically  or  mechanically  from 
an  outside  source.  Therefore,  the  iron  cores  used  for  variable  fluxes 
are  usually  built  of  laminations,  so  as  to  limit  the  eddy  currents 
to  a  small  amount  by  interposing  in  their  paths  the  insulation 
between  the  laminations.  Japan,  varnish,  tissue  paper,  etc.,  are 
used  for  this  purpose.  In  many  cases  the  layer  of  oxide  formed 
on  laminations  during  the  process  of  annealing  is  considered  to  be 
a  sufficient  insulation  against  eddy  currents. 

The  usual  thickness  of  lamination  varies  from  0.7  to  0.3  mm., 
according  to  the  frequency  for  which  an  apparatus  is  designed, 
the  flux  density  to  be  used,  the  provision  for  cooling,  etc.  The 
more  a  core  is  subdivided  the  lower  is  the  loss  due  to  eddy  currents, 
but  the  more  expensive  is  the  core  on  account  of  the  higher  cost 
of  rolling  sheets,  and  of  punching  and  assembling  the  laminations. 
Besides,  more  space  is  taken  by  insulation  with  thinner  stampings, 
so  that  the  per  cent  net  cross-section  of  iron  is  reduced.  The 
net  cross-section  of  laminations  is  usually  from  95  to  85  per  cent 
of  the  gross  cross-section,  depending  upon  the  thickness  of  the 
laminations,  the  kind  of  insulation  used,  and  the  care  and  pres- 
sure used  in  assembling  the  core.  For  preliminary  calculations 
about  ten  per  cent  of  the  gross  cross-section  is  assumed  to  be 
lost  in  insulation. 

Fig.  9  shows  two  iron  cores  in  cross-section,  one  core  solid, 
the  other  subdivided  into  three  laminations  by  planes  parallel 
to  the  direction  of  the  lines  of  force.  The  lines  of  force  are  shown 
by  dots,  and  the  paths  of  the  eddy  currents  by  continuous  lines. 
Eddy  currents  are  linked  with  the  lines  of  force,  the  same  as  the 
current  in  the  exciting  winding.  In  fact,  eddy  currents  are  similar 
to  the  secondary  currents  in  a  transformer,  inasmuch  as  they 
tend  to  reduce  the  flux  created  by  the  primary  current.  The 
core  must  be  laminated  in  planes  perpendicular  to  the  lines  of 
flow  of  the  eddy  currents,  so  as  to  break  up  their  paths  and  at  the 
same  time  not  to  interpose  air-gaps  in  the  paths  of  the  lines  of 
force. 

An  iron  core  can  be  further  subdivided  by  using  thin  iron 
wires  in  place  of  laminations.  Such  cores  were  used  in  early 
machines  and  transformers,  but  were  abandoned  on  account  of 


42 


THE  MAGNETIC  CIRCUIT 


[ART.  18 


expense  and  poor  space  factor.  Iron-wire  cores  are  used  at  present 
in  only  high-frequency  apparatus,  in  which  eddy  currents  must 
be  carefully  guarded  against;  for  instance  in  the  induction  coils 
(transformers)  employed  in  telephone  circuits. 

It  will  be  seen  by  an  inspection  of  Fig.  9  that  eddy  currents 
are  much  smaller  in  the  laminated  core  because  the  resistance  of 
each  lamination  is  increased  while  the  flux  per  lamination  and  con- 
sequently the  induced  e.m.f .  is  considerably  reduced.  It  is  proved 
in  Art.  21  below  that  the  power  lost  in  eddy  currents  per  kilogram 
of  laminations  is  proportional  to  the  square  of  the  thickness  of 


FIG.  9. — Eddy  currents  in  a  solid  and  in  a  laminated  core. 

the  laminations,  the  square  of  the  frequency,  and  the  square  of 
the  flux  density. 

Prob.  8.  Show  that  the  armature  cores  of  revolving  machinery  must 
be  laminated  in  planes  perpendicular  to  the  axis  of  rotation. 

Prob.  9.  Show  that  assuming  the  temperature-resistance  coefficient  of 
iron  laminations  to  be  0.0046  per  degree  Centigrade  the  eddy  current  loss  of 
a  core  at  70°  C.  is  only  about  82.5  per  cent  of  that  at  20°  C. 

Prob.  10.  Explain  the  reason  for  which  the  hysteresis  loss  in  a  given 
core  and  at  a  given  frequency  depends  only  on  the  amplitude  of  the  excit- 
ing current,  while  the  eddy-current  loss  depends  also  upon  the  wave-form 
of  the  current. 

18.  The  Significance  of  Iron  Loss  in  Electrical  Machinery. 

The  power  lost  in  an  iron  core  on  account  of  hysteresis  and  eddy 
currents,  taken  together,  is  called  iron  loss  or  core  loss.  It  is  of 
importance  to  understand  the  effect  of  this  loss  in  the  iron  cores 


CHAP.  Ill]  HYSTERESIS  AND  EDDY  CURRENTS  43 

of  electrical  machinery  and  apparatus:  First,  because  they  bring 
about  a  loss  of  power  and  hence  lower  the  efficiency  of  a 
machine;  Secondly,  because  they  heat  up  the  iron  and  thus 
limit  the  permissible  flux  density,  or  make  extra  provisions  for 
ventilation  and  cooling  necessary;  Thirdly,  because  they  affect 
the  indications  of  measuring  instruments.  The  effects  of  hystere- 
sis and  eddy  currents  in  the  principal  types  of  electrical  machinery 
are  as  follows: 

(a)  In  a  transformer  an  alternating  magnetization  of  the  iron 
causes  a  core  loss  in  it.     The  power  thus  lost  must  be  supplied  from 
the  generating  station  in  the  form   of  an  additional  energy  com- 
ponent of  the  primary  current.     The  core  is  heated  by  hysteresis 
and  by  eddy  currents,  and   the  heat  must  be  dissipated  by  the 
oil  in  which  the  transformer  is  immersed,  or  by  an  air  blast. 

(b)  In  a  direct-current  machine  the  revolving  armature  is 
subjected  to  a  magnetization  first  in  one  direction  and  then  in 
the  other;  the  heating  effect  due  to  the  hysteresis  and  eddy 
currents  is  particularly  noticeable  in  the  armature  teeth  in  which 
the  flux  density  is  usually  quite  high.     The  core  loss,  being  sup- 
plied mechanically,  causes  an  additional  resisting  torque  between 
the  armature  and  the  field.     In  a  generator  this  torque  is  sup- 
plied by  the  prime  mover;  in  a  motor  this  torque  reduces  the 
available  torque  on  the  shaft. 

(c)  The  effect  of  hysteresis  and  of  eddy  currents  in  the  armature 
of  an  alternator  or  of  a  synchronous  motor  is  similar  to  that  in  a 
direct-current  machine. 

(d)  In  an  induction  motor  the  core  loss  takes  place  chiefly 
in  the  stator  iron  and  teeth,  where  the  frequency  of  the  magnetic 
cycles  is  equal  to  that  of  the  power  supply;  the  frequency  in  the 
rotor  corresponds  to  the  per  cent  slip,  so  that  even  with  very 
high  flux  densities  in  the  rotor  teeth  the  core  loss  in  the  rotor  is 
comparatively  small.     At  speeds  below  synchronism  the  necessary 
power  for  supplying  the  iron  loss  is  furnished  electrically  as  part 
of  the  input  into  the  stator.     At  speeds  above  synchronism  this 
power  is  supplied  through  the  rotor  from  the  prime  mover. 

(e)  In  a  direct-current  ammeter,  if  it  has  a  piece  of  iron  as 
its  moving  element,  residual  magnetism  in  this  iron  causes  inac- 
curacies in  its  indications.     With  the  same  current  the  indication 
of  the  instrument  is  smaller  when  the  current  is  increasing  than 
when  it  is  decreasing;  this  can  be  understood  with  reference  to 


44  THE  MAGNETIC  CIRCUIT  [ART.  19 

the  hysteresis  loop.  With  alternating  current  the  effect  of  hystere- 
sis is  automatically  eliminated  by  the  reversals  of  the  current 
which  passes  through  the  instrument. 

From  these  examples  the  reader  can  judge  as  to  the  effect  of 
hysteresis  in  other  types  of  electrical  apparatus  not  considered 
above. 

Prob.  11.  Show  that  in  an  8-pole  direct-current  motor  running  at  a 
speed  of  525  r.p.m.  the  armature  core  and  teeth  undergo  35  complete 
hysteresis  cycles  per  second. 

Prob.  12.  Show  that  for  two  points  in  an  armature  stamping,  taken 
on  the  same  radius,  one  in  a  tooth,  the  other  near  the  inner  periphery  of 
the  armature,  the  hysteresis  loops  are  displaced  in  time  by  one-quarter 
of  a  cycle. 

19.  The  Total  Core  Loss.  In  practical  calculations  on  electrical 
machinery  the  total  core  loss  is  of  interest,  rather  than  the  hystere- 
sis and  the  eddy  current  losses  separately.  For  such  computations 
empirical  curves  are  used,  obtained  from  tests  on  steel  of  the  same 
quality  and  thickness.  The  curves  of  total  core  loss  given  in 
Fig.  10  have  been  compiled  from  various  sources,  and  give  a 
fak  idea  of  the  order  of  magnitude  of  core  loss  in  various  grades 
of  commercial  steel  laminations.  The  specimens  were  tested  in 
the  Epstein  apparatus,  which  is  a  miniature  transformer  (see 
the  author's  Experimental  Electrical  Engineering,  Vol.  1,  p.  197), 
and  the  values  given  can  be  used  for  estimating  the  core  loss  in 
transformers  and  in  other  stationary  apparatus  with  a  simple 
magnetic  circuit. 

In  using  the  curves  one  should  note  that  the  ordinates  are  watts 
per  cubic  decimeter  of  laminations,  hence  the  gross  volume  and 
not  the  volume  of  the  iron  itself  is  represented.  On  the  other  hand, 
the  abscissae  are  the  true  flux  densities  in  the  iron.  In  choosing  a 
material  the  following  points  are  worthy  of  note :  (1)  Silicon  steel 
is  now  used  for  60-cycle  transformers,  almost  to  the  exclusion  of 
any  other,  on  account  of  its  lower  core  loss ;  it  is  sometimes  used 
for  25-cycle  transformers  also.  (2)  The  material  called  "  Good 
carbon  steel "  is  that  which  is  used  for  induction  motor  stators, 
and  in  general  for  the  armatures  of  alternating  and  direct- 
current  machinery ;  also,  sometimes  for  the  cores  oHow  frequency 
transformers.  (3)  The  material  called  "  Ordinary  carbon  steel  " 
should  be  used  only  in  those  cases  for  which  the  core  loss  is  of 
small  importance. 


CHAP.  Ill] 


HYSTERESIS  AND  EDDY  CURRENTS 


45 


46  THE  MAGNETIC  CIRCUIT  [ART.  19 

The  thickness  of  lamination  to  be  used  in  each  particular 
case  is  a  matter  of  judgment  based  on  previous  experience,  and 
no  general  rule  can  be  laid  down,  except  what  is  said  in  Art.  17 
above,  in  regard  to  the  factors  upon  which  the  eddy -current  loss 
depends.  The  gauges  26  to  29  are  representative  of  the  usual 
practice.  If  it  should  be  necessary  to  estimate  the  core  loss  for  a 
different  thickness  and  at  another  frequency  than  those  given  in 
Fig.  10,  the  method  explained  in  Art.  22  below  may  be  used. 

The  core  loss  in  the  armatures  and  teeth  of  revolving  machinery 
is  found  from  tests  to  be  considerably  above  that  calculated  from 
the  curves  of  loss  on  the  same  material  when  tested  in  stationary 
strips.  This  is  probably  due  in  part  to  the  fact  that  the  conditions 
of  magnetization  are  different  in  the  two  cases.  In  the  one  case 
the  cycles  of  magnetization  are  due  to  a  pulsating  m.m.f.,  which 
simply  changes  its  magnitude;  in  the  other  case  to  a  gliding  m.m.f., 
with  which  the  magnetic  intensity  at  a  point  changes  its  direction 
as  well.  Besides,  the  distribution  of  flux  densities  in  teeth  and  in 
armature  cores  is  very  far  from  being  uniform.  Therefore, 
when  using  the  curves  given  in  Fig.  10,  for  the  calculation  of  iron 
loss  in  generators  and  motors,  it  is  necessary  to  multiply  the  results 
by  certain  empirical  coefficients  obtained  from  the  results  of  tests 
made  on  similar  machines.  Mr.  I.  E.  Hanssen  recommends  add- 
ing 30,  35,  and  40  per  cent  to  the  loss  calculated  from  the  curves 
obtained  on  stationary  samples  when  estimating  the  iron  loss  in 
an  armature  back  of  its  teeth,  at  25,  40,  and  60  cycles  respectively. 
For  teeth  he  recommends  adding  30,  60,  and  80  per  cent,  at  the 
same  frequencies.1  These  values  are  quoted  here  merely  to  give 
a  general  idea  of  the  magnitude  of  the  excess  of  core  loss  in  revolv- 
ing machinery;  a  responsible  designer  should  compile  the  values 
of  such  coefficients  from  actual  tests  made  on  the  particular  class 
of  machines  which  he  is  designing. 

Some  engineers  do  not  use  for  revolving  machinery  values  of 
core  loss  obtained  on  stationary  samples, but  plot  the  curves  of  core 
loss  obtained  directly  from  tests  on  machines  of  a  particular  kind, 
for  various  frequencies  and  flux  densities.  This  is  a  reliable  and 
convenient  method  provided  that  sufficient  data  are  available  to 
separate  the  core  loss  in  the  teeth  from  that  in  the  core  itself.  Mr. 
H.  M.  Hobart  advocates  this  method,  and  curves  of  core  loss 

1  Hanssen,  "  Calculation  of  Iron  Losses  in  Dynamo-electric  Machinery," 
Trans.  Amer.  Inst.  Elec.  Eng.,  Vol.  28  (1909),  Part  II,  p.  993. 


CHAP.  Ill]  HYSTERESIS  AND  EDDY  CURRENTS  47 

obtained  directly  from  actual  machines  will  be  found  in  his  several 
books  on  electric  machine  design. 

It  is  customary  now  to  characterize  a  lot  of  steel  laminations 
with  respect  to  its  core  loss  by  the  co-called  figure  of  loss  (Verlust- 
ziffer) ,  which  is  the  total  core  loss  in  watts  per  unit  of  weight,  at  a 
standard  frequency  and  flux  density.  In  Europe  the  figure  of  loss 
is  understood  to  be  the  watts  loss  per  kilogram  of  laminations,  at 
50  cycles  and  at  a  flux  density  of  10  kilolines  per  square  centimeter; 
the  test  to  be  performed  in  an  Epstein  apparatus  under  definitely 
prescribed  conditions.1  Sometimes  a  second  figure  of  loss  is 
required,  referring  to  a  density  of  15  kilolines  per  square  centi- 
meter, when  the  laminations  are  to  be  used  at  high  flux  densities. 
In  this  country  a  figure  of  loss  is  sometimes  used  which  gives  the 
watts  loss  per  pound  of  material  at  60  cycles  and  at  a  flux  density 
of  60  kilolines  per  square  inch  (or  else  at  10  kilolines  per  square 
centimeter;  see  the  paper  mentioned  in  problem  20  below). 

In  some  cases  it  is  required  to  estimate  the  hysteresis  and  the 
eddy  current  losses  separately ;  also  it  is  sometimes  necessary  to 
separate  the  two  losses  knowing  a  curve  of  the  total  loss.  These 
calculations  are  explained  in  the  articles  that  follow. 

Prob.  13.  The  core  of  a  60-cycle  transformer  weighs  89  kg.;  the 
gross  cross-section  of  the  core  is  8  by  10  cm.,  of  which  10  per  cent  is  taken 
by  the  insulation  between  the  laminations.  The  total  flux  alternates 
between  the  values  of  ±0.49  megaline.  If  the  core  is  made  of  gauge  26 
good  carbon  steel,  what  is  the  total  core  loss  according  to  the  curves  in 
Fig.  10?  Solution:  The  flux  density  is  490/(8X  10X0.9)  =6.8  kl/sq.  cm. 
The  core  loss  per  cubic  decimeter  at  this  density  and  at  60  cycles  is, 
according  to  the  curve,  equal  11.5  watt.  The  volume  of  the  laminations, 
including  the  insulation,  is  89/7  =  12.7  cu.  dm.  The  total  loss  is  11.5X 
12.7  =  146  watts.  Ans.  146  watts. 

Prob.  14.  What  flux  density  could  be  used  in  the  preceding  problem 
if  the  core  were  made  of  silicon-steel  laminations,  gauge  29,  provided  that 
the  total  core  loss  be  kept  the  same  in  both  cases? 

Ans.    About  9  kl/sq.  cm. 

Prob.  15.  Calculate  the  core  loss  in  the  stationary  armature  of  a  60- 
cycle  450-r.p.m.  alternator  of  the  following  dimensions:  bore  180  cm.; 
gross  axial  length  24  cm.;  two  air-ducts  0.8  cm.  each;  radial  width  of 
stampings  back  of  the  teeth,  15  cm.;  the  machine  has  144  slots,  2  cm. 
wide  and  4.5  cm.  deep.  The  core  is  made  of  26-gauge  good  carbon  steel; 
the  useful  flux  per  pole  is  4.65  megalines,  and  two-thirds  of  the  total  num- 
ber of  teeth  carry  the  flux  simultaneously.  Use  Mr.  Hanssen's  coefficients. 

1  See  Ekktrotechnische  Zeitschrift,  Vol.  24  (1903),  p.  684. 


48  THE  MAGNETIC  CIRCUIT  [AKT.  20 

Note:  All  parts  of  the  core  and  all  the  teeth  are  subjected  to  complete 
cycles  of  magnetization  in  succession  ;  therefore,  in  calculating  the  core 
loss  the  total  volume  of  the  core  and  of  the  teeth  must  be  multiplied  by 
the  loss  per  cubic  decimeter,  corresponding  to  the  maximum  magnetic 
density  in  each  part.  "The  density  in  a  tooth  varies  along  its  length, 
being  a  maximum  at  the  tip.  The  average  density  may  be  assumed  to  be 
equal  to  that  at  the  middle  of  the  teeth.  Ans.  About  9  kw. 

•20.  Practical  Data  on  Hysteresis  Loss.  The  energy  lost  in 
hysteresis  per  cycle  per  kilogram  of  a  given  material  depends  only 
upon  the  maximum  values  of  B  and  H}  and  does  not  depend  upon 
the  manner  in  which  the  magnetizing  current  is  varied  with  the  time 
between  its  positive  and  negative  maxima.  It  is  only  at  very  high 
frequencies,  such  as  are  used  in  wireless  telegraphy,  that  the  par- 
ticles of  iron  do  not  seem  to  be  able  to  follow  in  their  grouping  the 
corresponding  changes  in  the  exciting  current.  With  such  high 
frequencies  iron  cores  are  not  only  useless,  but  positively  harmful. 
However,  at  ordinary  commercial  frequencies  the  loss  of  power 
Ph  due  to  hysteresis  is  proportional  to  the  number  of  cycles  per 
second  and  can  be  expressed  as 

Ph=f'V-F(B)  watt., 

where  /  is  the  number  of  magnetic  cycles  per  second,  V  is  the  vol- 
ume of  the  iron,  and  F(B)  is  a  function  of  the  maximum  flux  den- 
sity B.  F(B)  represents  the  loss  per  cycle  per  cubic  unit  of 
material,  and  is  therefore  equal  to  the  area  of  the  hysteresis  loop 
in  Fig.  7. 

One  can  assume  empirically  that  the  unit  loss  per  cycle,  F(B), 
increases  as  a  certain  power  n  of  B}  this  power  to  be  determined 
from  tests.  The  preceding  formula  becomes  then 


(20) 


where  ^  is  an  empirical  coefficient  which  depends  upon  the  quality 
of  the  iron  and  upon  the  units  used.  Dr.  Steinmetz  found  from 
numerous  experiments  that  the  exponent  n  varies  between  1.5  and 
1.7,  and  proposed  for  practical  use  the  formula 

ph=  77/7^1  -e  x  10~7  watt,       ....     (21) 


where  the  factor  10~7  is  introduced  in  order  to  obtain  convenient 
values  for  y  when  B  is  in  maxwells  per  square  centimeter,  and  V 
is  in  cubic  centimeters.  It  is  more  convenient  for  practical  calcula- 


CHAP.  Ill]  HYSTERESIS  AND  EDDY  CURRENTS  49 

tions  to  use  B  in  kilolines  per  square  centimeter,  and  V  in  cubic 
decimeters.  In  this  case  the  constant  10~7  is  not  necessary  (see 
Prob.  17  below)  ;  but  the  student  must  now  remember  to  multiply 
by  6.31  the  values  of  T?  found  in  the  various  pocketbooks. 

Hysteresis  loss  cannot  be  represented  always  with  sufficient 
accuracy  by  formula  (21)  or  (20)  over  a  wide  range  of  values  of  B, 
because  the  exponent  n  itself  seems  to  increase  with  B.  Where 
greater  accuracy  is  required  at  medium  and  high  flux  densities  the 
following  formula  may  be  used: 

)  ......     (21a) 


In  this  formula  the  term  with  B2  automatically  becomes  of  more 
and  more  importance  as  B  increases.  By  selecting  proper  values 
for  r/  and  if'  a  given  experimental  curve  of  loss  can  be  approxi- 
mated more  closely  than  by  means  of  formula  (21).  On  the  other 
hand,  formula  (21)  is  more  convenient  for  comparison  and  analysis. 

Curves  of  hysteresis  loss  and  values  of  the  constant  rj  will  be 
found  in  various  handbooks  and  pocketbooks.  It  is  hardly  worth 
while  giving  them  here,  because  hysteresis  loss  varies  greatly  with 
the  quality  of  iron  and  with  the  treatment  it  is  given  before  use. 
Moreover,  the  quality  of  the  iron  used  in  electrical  machinery  is 
being  improved  all  the  time,  so  that  a  value  of  r)  given  now  may 
be  too  large  a  few  years  from  now. 

Considerable  effort  is  being  constantly  made  to  improve  the 
quality  of  the  iron  used  in  electrical  machinery  so  as  to  reduce  its 
hysteresis  loss.  The  latest  achievement  in  this  respect  is  the  pro- 
duction of  the  so-called  silicon  steel,  also  called  alloyed  steel,  which 
contains  from  2.5  to  4  per  cent  of  silicon.  This  steel  shows  a  much 
lower  hysteresis  loss  than  ordinary  carbon  steel.  Incidentally, 
the  electric  resistivity  of  silicon  steel  is  about  three  times  higher 
than  that  of  ordinary  steel,  so  that  the  eddy  -current  loss  is  reduced 
about  three  times.  The  advantage  that  silicon  steel  has  over  car- 
bon steel  is  clearly  seen  in  Fig.  10.  Silicon  steel  is  largely  used 
for  transformer  cores  because  it  permits  the  use  of  higher  flux 
densities,  and  therefore  the  reduction  of  the  weight  and  cost  of  a 
transformer,  in  spite  of  the  fact  that  silicon  steel  itself  costs  more 
per  kilogram  than  carbon  steel. 

Another  great  advantage  of  silicon  steel  is  that  it  is  practically 
non-aging-,  this  means  that  the  hysteresis  loss  does  not  increase 
with  time.  An  increase  in  the  hysteresis  loss  of  a  transformer 


50  THE  MAGNETIC  CIRCUIT  [ART.  20 

during  the  first  few  years  of  its  operation  used  to  be  a  serious 
matter  in  the  design  and  operation  of  transformers,  because  of  the 
subsequent  overheating  of  the  core  and  of  the  coils.  Silicon  steel 
shows  practically  no  increase  in  its  hysteresis  loss  after  several 
years  of  operation.  Moderate  heating,  which  considerably  in- 
creases the  hysteresis  loss  in  ordinary  steel,  has  no  effect  on  silicon 
steel. 

Impurities  which  are  of  such  a  nature  as  to  produce  a  softer 
iron  or  steel  and  a  material  of  higher  permeability,  are  as  a  rule 
favorable  to  the  reduction  of  the  hysteresis  loss,  and  vice  versa. 
Mechanical  treatment  and  heating  are  also  very  important  in  their 
effects  on  hysteresis  loss.  In  particular,  punching  and  hammering 
increases  hysteresis  loss,  while  annealing  reduces  it.  Therefore 
laminations  are  always  annealed  carefully  after  being  punched 
into  their  final  shape. 

The  requirements  for  the  steel  used  in  permanent  magnets  are 
entirely  different  from  those  for  the  cores  of  electrical  machinery. 
In  permanent  magnets  a  large  and  wide  hysteresis  loop  is  desired, 
because  it  means  a  high  percentage  of  residual  magnetism  (ratio 
of  CO  to  AP,  Fig.  7)  and  a  large  coercive  force,  OF.  Both  are 
favorable  for  obtaining  strong  permanent  magnets  of  lasting 
strength.  Combined  carbon  is  particularly  important  for  obtain- 
ing these  qualities,  as  is  also  the  proper  heat  treatment  after  mag- 
netization. 

Prob.  16.  In  the  60-cycle  transformer  given  in  prob.  13,  the  core 
weighs  89  kg.  and  is  made  of  26  gauge  good  carbon  steel.  The  maxi- 
mum flux  density  is  6.8  kl./sq.  cm.  What  is  the  hysteresis  loss  assuming 
T?  to  be  equal  to  0.0012?  Ans.  About  124  watt. 

Prob.  17.  What  is  the  constant  in  formula  (21)  in  place  of  10~7,  if, 
with  the  same  >?,  the  density  B  is  in  kilo-maxwells  per  sq.  cm.,  and  the 
volume  is  in  cubic  decimeters?  Ans.  6.31. 

Prob.  18.  Show  how  to  determine  the  values  of  y  and  nin  eq.  (20), 
knowing  the  values  W i  and  W2  of  the  energy  lost  per  cycle  at  two  given 
values  of  maximum  flux  density,  J5t  and  B2. 

Ans.  n  =  (log  TF2-log  WJ/(log  B,— log  BJ. 

Prob.  19.  The  following  values  of  hysteresis  loss  per  cu.  decimeter 
have  been  determined  from  a  test  at  25  cycles  (after  eliminating  the  eddy 
current  loss) : 


Flux  density  in  kl/sq.cm.,  B  =  5. 0    I  6 . 5 


Hysteresis  loss  in  watts,    Ph  =  1 . 30  I  2 . 00 


8.0 

2.8* 


10.0 
4.11 


What  are  the  values  of  TJ  and  n  in  formula  (20)?     Suggestion; 
Use  logarithmic  paper  to  determine  the  most  probable  value  of  n,  by 


CHAP.  Ill]  HYSTERESIS  AND  EDDY  CURRENTS  51 

drawing  the  straight  line  log  P/i=n  log  B+  log  Const.  See  the  author's 
Experimental  Electrical  Engineering,  Vol.,  1,  p.  202. 

Ans.    Ph  =  0.00368  fVB1'™. 

21.  Eddy  Current  Loss  in  Iron.  With  the  thin  laminations 
used  in  the  cores  of  electrical  machinery  the  eddy -current  loss  in 
watts  can  be  represented  by  the  formula 

Pe=£V(tfB)2,  .     .     /..'.     .     .     (22) 

where  e  is  a  constant  which  depends  upon  the  electrical  resistivity 
of  the  iron,  its  temperature,  the  distribution  of  the  flux,  the  wave 
form  of  the  exciting  current,  and  the  units  used.  V  is  the  volume 
or  the  weight  of  the  core  for  which  the  loss  is  to  be  computed ;  t  is 
the  thickness  of  laminations,  / the  frequency  of  the  supply,  and  B 
the  maximum  flux  density  during  a  cycle.  If  B  is  different  at 
different  places  in  the  same  core,  the  average  of  these  should 
be  taken,  (B  is  the  time  maximum  but  the  space  average). 
Sometimes  formula  (22)  contains  also  10  to  some  negative  power 
in  order  to  obtain  a  convenient  value  of  e. 

Formula  (22)  can  be  proved  as  follows :  The  loss  of  power  in  a 
lamination  can  be  represented  as  a  sum  of  the  i2r  losses  for  the 
small  filaments  of  eddy  current  in  it.  But  i2r=e2/r-}  it  can  be 
shown  that  the  expression  in  parentheses  in  formula  (22)  is  pro- 
portional to  the  sum  of  e2/r  per  unit  volume.  When  the  frequency 
/  increases  say  n  times,  the  rate  of  change  of  the  flux,  dQ/dt,  and 
consequently  the  e.m.fs.  induced  in  the  iron  are  also  increased  n 
times.  Therefore,  the  loss  which  is  proportional  to  e2  increases  n2 
times.  In  other  words,  the  loss  is  proportional  to  the  square  of 
the  frequency.  Similarly,  the  induced  voltage  is  proportional  to 
the  flux  density  J5;  and  consequently,  the  loss  is  proportional  to  B2. 

To  prove  that  the  loss  is  proportional  to  the  square  of  the 
thickness  of  laminations  one  must  remember  that  increasing  the 
thickness  n  times  increases  the  flux  and  the  induced  e.m.f .  within 
any  filament  of  eddy  current  also  n  times.  But  the  resistance  of 
each  path  is  reduced  n  times  (neglecting  the  short  sides  of  the  rect- 
angle). Consequently,  the  expression  e2/r  is  increased  n3  times. 
However,  inasmuch  as  the  volume  of  the  lamination  is  also 
creased  n  times,  the  loss  per  unit  volume  is  only  n2  times  larger.  In 
other  words,  the  loss  per  unit  volume  increases  as  t2.  A  more 
rigid  proof  of  this  proposition  is  given  in  problem  21  below. 


52  THE  MAGNETIC  CIRCUIT  [ABT.  22 

For  values  of  e  the  reader  is  referred  to  pocketbooks;  the 
numerical  values  given  there  must,  however,  be  used  cautiously, 
because  the  eddy -current  loss  depends  on  some  factors  such  as  the 
care  exercised  in  assembling,  and  the  actual  distribution  of  the 
flux,  which  factors  can  hardly  be  taken  into  account  in  a  formula. 
As  a  matter  of  fact,  formula  (22)  is  used  now  less  and  less  in  prac- 
tical calculations,  the  engineer  relying  more  upon  experimental 
curves  of  total  core  loss  (Fig.  10). 

Prob.  20.  According  to  the  experiments  of  Lloyd  and  Fischer  [Trans- 
Amer.  Inst.  Eke.  Engs.,  Vol.  28  (1909),  p.  465]  the  eddy-current  loss  in 
silicon-steel  laminations  of  gauge  29  (0.357  mm.  thick)  is  from  0.12  to  0.18 
watts  per  pound  at  60  cycles  and  at  B  =  10,000  maxwells  per  sq.  cm.  What 
is  the  value  of  the  coefficient  £  in  formula  (22)  if  P  is  in  microwatts,  V  is 
the  weight  of  the  core  in  kg.  (not  the  volume,  as  before) ;  if  also  t  is  in  mm., 
and  B  is  in  kilolines  per  sq.  cm.? 

Ans.    From  5.78  to  8.67 ;  7.2  is  a  good  practical  average. 

Prob.  21.  Prove  that  the  loss  of  power  caused  by  eddy  currents,  per 
unit  volume  of  thin  laminations,  is  proportional  to  the  square  of  the  thick- 
ness of  the  laminations.  Solution :  The  thickness  t  of  the  sheet  (Fig.  9) 
being  by  assumption  very  small  as  compared  with  its  width  a,  the  paths 
of  the  eddy  current  may  be  considered  to  be  rectangles  of  the  length  a  and 
of  different  widths,  ranging  from  t  to  zero.  Consider  one  of  the  tubes  of 
flow  of  current,  of  a  width  2x,  thickness  dx,  and  length  I  in  the  direction 
of  the  lines  of  magnetic  force.  Let  the  flux  density  vary  with  the  time 
between  the  limits  ±B.  Then  the  maximum  flux  linking  with  the  tube 
of  current  under  consideration  is  approximately  equal  to  2axB ;  therefore, 
the  effective  value  of  the  voltage  induced  in  the  tube  can  be  written  in  the 
form  e  =  CaxBf,  where  C  is  a  constant,  the  value  of  which  we  are  not  con- 
cerned with  here.  The  resistance  of  the  tube  is  p(2a  +4x)/(ldx),  or  very 
nearly  2ap/(ldx).  Thus  we  have  that  the  i2r  loss,  or  the  value  of  e*/r 
for  the  tube  under  consideration,  is  dPe  =  C2ax2B2f2ldx/2p.  Integrat- 
ing this  expression  between  the  limits  0  and  t/2  we  get  Pe=  C2at3B'*/2l/  48p. 
But  the  volume  of  the  lamination  is  V=atl.  Dividing  P  by  V  we  find 
that  the  loss  per  unit  volume  is  proportional  to  (t/B}2.1 

Prob.  22.  Prove  that  the  loss  of  power  by  eddy  currents  per  unit 
volume  in  round  iron  wires  is  proportional  to  the  square  of  the  diameter  of 
the  wire.  The  flux  is  supposed  to  pulsate  in  the  direction  of  the  axes  of 
the  wires,  and  the  lines  of  flow  of  the  eddy  currents  are  concentric  circles 
Hint :  Use  the  method  employed  in  the  preceding  problem. 

22.  The  Separation  of  Hysteresis  from  Eddy  Currents.  It  is 
sometimes  required  to  estimate  the  total  core  loss  for  a  thickness 

^or  a  complete  solution  of  this  and  the  following  problem,  including 
the  numerical  values  of  C,  see  Steinmetz,  Alternating  Current  Phenomena 
(1908),  Chap.  XIV. 


CHAP.  IIIl  HYSTERESIS  AND  EDDY  CURRENTS  53 

of  steel  laminations  other  than  those  given  in  Fig.  10,  or  at  a  differ- 
ent frequency.  For  this  purpose,  it  is  necessary  to  separate  the 
loss  due  to  hysteresis  from  that  due  to  eddy  currents,  because  the 
two  losses  follow  different  laws,  expressed  by  eqs.  (20)  and  (22) 
respectively. 

In  order  to  separate  these  losses  at  a  certain  flux  density  it  is 
necessary  to  know  the  value  of  the  total  core  loss  at  this  density, 
and  at  two  different  frequencies.  For  a  given  sample  of  lamina- 
tions, the  total  core  loss  P  at  a  constant  flux  density  and  at  a 
variable  frequency  /,  can  be  represented  in  the  form 

P=Hf+Ff2,       ......     (23a) 

where  Hf  represents  the  hysteresis  loss,  and  Ff2  the  eddy  or  Fou- 
cault  current  loss.  H  is  the  hysteresis  loss  per  cycle,  and  F  is  the 
eddy  -current  loss  when  /is  equal  to  one  cycle  per  second.  Writing 
this  equation  for  two  known  frequencies,  two  simultaneous  equa- 
tions are  obtained  for  H  and  F,  from  which  H  and  F  can  be  deter- 
mined. 

In  practice  the  preceding  equation  is  usually  divided  by  /,  be- 
cause in  the  form 

P/f=H  +  Ff,    .......     (236) 


it  represents  the  equation  of  a  straight  line  between  P/f  and  /. 
This  form  is  particularly  convenient  when  the  values  of  P  are 
known  for  more  than  two  frequencies.  In  this  case  the  values  of 
P/f  are  plotted  against  /  as  abscissae,  and  the  most  probable 
straight  line  is  drawn  through  the  points  thus  obtained.  The 
intersection  of  this  straight  line  with  the  axis  of  ordinates  gives 
directly  the  value  of  H.  After  this,  F  is  found  from  eq.  (236). 

Knowing  H  and  F  at  a  certain  flux  density,  the  separate  losses 
Hf  and  Ff2  can  be  calculated  for  any  desired  frequency.  For  the 
same  material,  but  of  a  different  thickness,  the  hysteresis  loss  per 
kilogram  weight  is  the  same,  while  the  eddy  -current  constant  F 
varies  as  the  square  of  the  thickness,  according  to  eq.  (22).  Thus, 
knowing  the  eddy  loss  at  one  thickness  it  can  be  estimated  for  any 
other  thickness. 

It  is  sometimes  required  to  estimate  the  iron  loss  at  a  flux  den- 
sity higher  than  the  range  of  the  available  curves  ;  in  other  words, 
the  problem  is  sometimes  put  to  extrapolate  a  curve  like  one  of 
those  in  Fig.  10.  There  are  two  cases  to  be  considered. 


54  THE  MAGNETIC  CIRCUIT  [ART.  22 

(A)  If  two  or  more  curves  for  the  same  material  are  available, 
taken  at  different  frequencies,  the  hysteresis  is  first  separated  from 
the  eddy  -current  loss  as  is  explained  before,  for  several  flux  densi- 
ties within  the  range  of  the  curves.     Then  the  exponent,  according 
to  which  the  hysteresis  loss  varies  with  the  flux  density  is  found, 
by  plotting  the  hysteresis  loss  to  a  logarithmic  scale  (see  problems 
18  and  19  above).     Finally  the  two  losses  are  extrapolated.     In 
extrapolating,  the  hysteresis  loss  is  assumed  to  vary  according  to 
the  same  law,  and  the  eddy  current  loss  is  assumed  to  vary  as  the 
square  of  the  flux  density;  see  eq.  (22). 

(B)  Should  only  one  curve  of  the  total  loss  be  available  for 
extrapolation,  this  curve  may  be  assumed  to  be  a  parabola  of  the 
form  P=aB  +  bB2.     Dividing  the  equation  throughout  by  B  we 
get 

(24) 


This  is  the  equation  of  a  straight  line  between  P/B  and  B.  Plot- 
ting P/B  against  the  values  of  B  as  abscissae,  a  straight  line  is 
obtained  which  can  be  easily  extrapolated.  In  some  cases  the 
values  of  P/B  thus  plotted  give  a  line  with  a  perceptible  curva- 
ture. Nevertheless,  the  curvature  is  much  smaller  than  that  of 
the  original  P  curve,  so  that  the  P/B  curve  can  be  extrapolated 
with  more  certainty,  especially  if  the  lower  points  be  disregarded.1 

Prob.  23.  From  the  curves  in  Fig.  10  calculate  the  core  loss  per  cubic 
decimeter  of  29-gauge  silicon-steel  laminations,  at  a  flux  density  of  10 
kl/sq.cm.  and  at  40  cycles.  Ans.  About  10  watts. 

Prob.  24.  Using  the  data  obtained  in  the  solution  of  the  preceding 
problem  calculate  the  figure  of  loss  of  26-gauge  laminations  at  60  cycles. 

Ans.  2.7  watt  /kg. 

Prob.  25.  Check  the  curve  of  total  core  loss  for  the  ordinary  carbon 
steel  at  40  cycles  with  the  curves  for  25  and  60  cycles. 

Prob.  26.  Extrapolate  the  curve  of  core  loss  for  the  silicon  steel  at  25 
cycles  up  to  the  density  of  20  kl/sq.cm.  Which  method  is  the  more 
preferable?  Ans.  22  watts  per  cu.dm.  at  B  =  20. 

Prob.  27.  Show  that  the  core  loss  curve  for  ordinary  carbon  steel,  at 
60  cycles,  follows  closely  eq.  (24). 

1  If  the  P/B  curve  should  prove  to  be  a  straight  line,  then  it  is  probable 
that  the  hysteresis  loss  follows  eq.  2  la  more  nearly  than  eq.  20.  In  this  case, 
even  if  we  had  data  for  two  frequencies,  method  (B)  would  be  both  more 
accurate,  and  more  simple  than  method  (A). 


CHAPTER  IV 
INDUCED  E.M.F.  IN  ELECTRICAL  MACHINERY 

23.  Methods  of  Inducing  E.M.F.  The  following  are  the  prin- 
cipal cases  of  induced  e.m.f.  in  electrical  machinery  and  apparatus: 

(a)  In  a  transformer,  an  alternating  magnetizing  current  in  the 
primary  winding  produces  an  alternating  flux  which  links  with 
both  windings  and  induces  in  them  alternating  e.m.fs.  A  similar 
case  is  that  of  a  variable  current  in  a  transmission  line  which 
induces  a  voltage  in  a  telephone  line  which  runs  parallel  to  it. 

(6)  In  a  direct-current  machine,  in  a  rotary  converter,  and  in 
a  homopolar  machine  electromotive  forces  are  induced  in  the 
armature  conductors  by  moving  them  across  a  stationary  magnetic 
field. 

(c)  In  an  alternator  and  in  a  synchronous  motor,  with  a  sta- 
tionary armature  and  a  revolving  field,  electromotive  forces  are 
induced  by  making  the  magnetic  flux  travel  past  the  armature 
conductors. 

(d)  In  a  polyphase  induction  motor  the  currents  in  the  stator 
and  in  the  rotor  produce  together  a  resultant  magnetomotive  force 
which  moves  along  the  air-gap  and  excites  a  gliding  (revolving) 
flux.     This  flux  induces  voltages  in  both  the  primary  and  the  sec- 
ondary windings. 

(e)  In  a  single-phase  motor,  with  or  without  a  commutator,  the 
e.m.fs.  induced  in  the  armature  are  partly  due  to  the  "  transformer 
action,"  as  under  (a),  and  partly  to  the  "  generator  action,"  as 
under  (6). 

(/)  In  an  inductor-type  alternator  both  the  exciting  and  the 
armature  windings  are  stationary;  the  pole  pieces  alone  revolve. 
The  flux  linked  with  the  armature  coils  varies  periodically,  due  to 
the  varying  reluctance  of  the  magnetic  circuit,  because  of  the 
motion  of  the  pole  pieces.  This  varying  flux  induces  an  alternating 
e.m.f.  in  the  armature  winding.  Or  else,  one  may  say  that  the 

55 


56  THE  MAGNETIC  CIRCUIT  [ART.  23 

flux  travels  along  the  air-gap  with  the  projecting  poles,  and  cuts 
the  armature  conductors. 

(g)'  Whenever  the  current  varies  in  a  conductor,  e.m.fs.  are 
induced  not  only  in  surrounding  conductors  but  also  in  the  con- 
ductor itself.  This  e.m.f.  is  called  the  e.m.f.  of  self-induction. 
Such  e.m.fs.  are  present  in  alternating-current  transmission  lines, 
in  the  armature  windings  of  alternating-current  machinery,  etc. 
While  the  e.m.f.  of  self-induction  does  not  differ  fundamentally 
from  the  transformer  action  mentioned  above,  its  practical  aspect 
is  such  as  to  make  a  somewhat  different  treatment  desirable. 
Inductance  and  its  effects  are  therefore  considered  separately  in 
chapters  X  to  XII  below. 

All  of  the  foregoing  cases  can  be  reduced  to  the  following  two 
fundamental  modes  of  action  of  a  magnetic  flux  upon  an  electrical 
conductor: 

(1)  The  exciting  magnetomotive  force  and  the  winding  in  which 
an  e.m.f.  is  to  be  induced  are  both  stationary,  relatively  to  one 
another;  in  this  case  the  voltage  is  induced  by  a  varying  magnetic 
flux.     Changes  in  the  flux  are  produced  by  varying  either  the 
magnitude  of  the  m.m.f .,  or  the  reluctance  of  the  magnetic  circuit. 
This  method  of  inducing  an  e.m.f.  is  usually  called  the  transformer 
action. 

(2)  The  exciting  magnetomotive  force  and  the  winding  in  which 
the  e.m.f.  is  to  be  induced  are  made  to  move  relatively  to  each 
other,  so  that  the  armature  conductors  cut  across  the  lines  of  the 
flux.     This  method  of  inducing  an  e.m.f.  is  conventionally  referred 
to  as  the  generator  action. 

By  analyzing  the  transformer  action  more  closely  it  can  be 
reduced  to  the  generator  action,  that  is  to  say  to  the  "  cutting  "  of 
the  secondary  conductor  by  lines  of  magnetic  flux  or  force.  This  is 
so,  because  in  reality  the  magnetic  disturbance  spreads  out  in  all 
directions  from  the  exciting  winding,  and  when  the  current  in  the 
exciting  winding  varies  the  magnetic  disturbance  travels  to  or 
from  the  winding  in  directions  perpendicular  to  the  lines  of  force 
(Fig.  11).  This  traveling  flux  cuts  the  secondary  conductor  and 
induces  in  it  an  e.m.f.  However,  the  question  as  to  whether  an 
e.m.f.  is  induced  by  a  change  in  the  total  flux  within  a  loop,  or  by 
the  cutting  of  a  conductor  by  a  magnetic  flux  is  still  in  a  somewhat 
controversial  state;1  although  Bering's  experiment  is  a  strong 

1  Carl  Hering,  "  An  Imperfection  in  the  Usual  Statement  of  the  Funda- 


CHAP.  IV] 


INDUCED  E.M.F. 


57 


argument  in  favor  of  the  theory  of  "  cutting  "  of  lines  of  force. 
He  showed  that  no  e.m.f.  is  induced  in  an  electric  circuit  when  a 
flux  is  brought  in  or  out  of  it  without  actually  cutting  any  of  the 
conductors  of  the  electric  circuit.  For  practical  purposes  it  is 
convenient  to  distinguish  the  transformer  action  from  the  genera- 
tor action,  so  that  the  matter  of  unifying  the  statements  (1)  and 
(2)  into  one  more  general  law  is  of  no  immediate  importance. 

24.  The  Formulae  for  Induced  E.M.F.      In  accordance  with  the 
definition  of  the  weber  given  in  Art.  3;  we  have 

e=-d$/dt, (25) 

where  e  is  the  instantaneous  e.m.f.  in  volts,  induced  by  the  trans- 


FIG.  11. — E.M.F.  induced  by  transformer  action. 

former  action  in  a  turn  of  wire  which  at  the  time  t  is  linked  with  a 
flux  of  fl>  webers.  The  value  of  e  is  determined  not  by  the  value  of 
$  but  by  the  rate  at  which  0  varies  with  the  time.  In  the  case  of  the 
generator  action  d®  in  formula  (25)  represents  the  flux  which  the 
conductor  under  consideration  cuts  during  the  interval  of  time  dt. 
It  can  be  shown  that  the  two  interpretations  of  dd>  lead  to  the 
same  result.  Namely, in  the  case  of  the  transformer  action  (Fig.  11), 
the  new  flux,  d@,  is  brought  within  the  secondary  turn  by  cutting 
through  the  conductor  of  this  turn.  Therefore,  in  the  case  of  the 

mental  Law  of  Electromagnetic  Induction,  Trans.  Amer.  Inst.  Elec.  Engs., 
Vol.  27  (1908),  Part.  2,  p.  .1341.  Fritz  Emde,  Dag  Induktionsgesetz,  Elek- 
trotechnik  und  Maschineribau,  Vol.  26  (1908);  Zum  Induktionsgesetz,  ibid., 
Vol.  27  (1909) ;  De  Baillehache,  Sur  la  Loi  de  1'Induction,  Bull.  Societe  Inter- 
nationale des  Ekctriciens,  Vol.  10  (1910),  pp.  89  and  288. 


58  THE  MAGNETIC  CIRCUIT  [ART.  24 

transformer  action  d@  can  also  be  considered  as  the  flux  which  cuts 
the  loop  during  the  time  dt,  the  same  as  in  the  generator  action. 
On  the  other  hand,  the  moving  conductor  in  a  generator  is  a  part 
of  a  turn  of  wire,  and  any  flux  which  it  cuts  either  increases  or 
decreases  the  total  flux  linking  with  the  loop.  Consequently,  in  the 
case  of  the  generator  action  d®  can  be  interpreted  as  the  change  of 
flux  within  the  loop,  the  same  as  in  the  transformer  action.  Thus, 
the  mathematical  expression  for  the  induced  e.m.f.  is  the  same 
in  both  cases,  provided  that  the  proper  interpretation  is  given  to 
the  value  of  d@. 

The  sign  minus  in  formula  (25)  is  understood  with  reference  to 
the  right-hand  screw  rule  (Art.  1),  i.e.,  with  reference  to  the  direc- 
tion of  the  current  which  would  flow  as  a  result  of  the  induced 
electromotive  force.  Namely,  the  law  of  the  conservation  of 
energy  requires  that  this  induced  current  must  oppose  any  change 
in  the  flux  linking  with  the  secondary  circuit.  If  this  were  other- 
wise, a  slight  increase  in  the  flux  would  result  in  a  further  indefinite 
increase  in  the  flux,  and  any  slight  motion  of  a  conductor  across  a 
magnetic  field  would  help  further  motion. 

The  positive  direction  of  the  induced  e.m.f.  is  understood  to  be 
that  of  the  primary  current  which  excites  the  flux  at  the  moment 
under  consideration.  If  the  flux  linked  with  the  secondary  circuit 
increases,  d@/dt  in  formula  (25)  is  positive,  but  the  secondary 
current  must  be  opposite  to  the  primary  in  order  to  oppose  the 
increase.  Thus,  the  secondary  current  is  negative,  and  by  assump- 
tion the  induced  e.m.f.  e  is  also  negative.  Therefore,  the  sign 
minus  is  necessary  in  the  formula.  When  the  flux  decreases, 
d$/dt  is  negative,  but  the  secondary  current  is  positive,  because  it 
must  oppose  the  reduction  in  flux.  Hence,  in  order  to  make  e  a 
positive  quantity,  the  sign  minus  is  again  necessary. 

The  following  two  special  cases  of  formula  (25)  are  convenient 
in  applications.  Formula  (25)  gives  the  instantaneous  value  of 
the  induced  e.m.f.;  it  is  once  and  a  while  required  to  know  the 
average  e.m.f.  induced  during  a  finite  change  of  the  flux  from  fl>i 
to  tf>2-  By  definition,  the  average  e.m.f.  is 

* 

edt, 


where  h  is  the  initial  moment  and  t2  the  final  moment  of  the 
interval  of  time  during  which  the  change  in  the  flux  takes  place. 


CHAP.  IV]  INDUCED  E.M.F.  59 

Substituting  in  this  equation  the  value  of  e  from  (25) ,  and  integ- 
rating, we  get 

-*i) (26) 


This  shows  that  the  average  value  of  an  induced  e.m.f.  does 
not  depend  upon  the  law  according  to  which  the  flux  changes  with 
the  time,  and  is  simply  proportional  to  the  average  rate  of  change 
of  the  flux. 

As  another  special  form  of  eq.  (25)  consider  a  straight  con- 
ductor of  a  length  I  centimeters  moving  at  a  velocity  of  v  centi- 
meters per  second  across  a  uniform  magnetic  field  of  a  density  of  B 
webers  per  sq.  cm.  Let  B,  I,  and  v  be  in  three  mutually  perpendic- 
ular directions.  The  flux  dtf>  cut  by  the  conductor  during  an  infini- 
tesimal element  of  time  dt  is  equal  to  Blvdt.  Substituting  this 
value  into  eq.  (25)  we  get,  apart  from  the  sign  minus, 

e=Blv (27) 

Should  the  three  directions,  B,  I,  and  v,  be  not  perpendicular  to 
each  other,  I  in  eq.  (27)  is  understood  to  mean  the  projection  of 
the  actual  length  of  the  conductor,  perpendicular  to  the  field,  and 
v  is  the  component  of  the  velocity  normal  to  B  and  1.  Both  B  and 
v  may  vary  with  the  position  of  the  conductor,  in  which  case  eq. 
(27)  gives  the  value  of  the  instantaneous  voltage.  If,  at  a  cer- 
tain moment,  the  various  parts  of  the  conductor  cut  across  a  field 
of  different  density,  eq.  (27)  must  be  written  for  an  infinitesimal 
length  of  the  conductor,  thus:  de=Bv-dl,  and  integrated  over 
the  whole  length  of  the  conductor. 

Besides  the  rule  given  above,  the  direction  of  the  e.m.f.  induced 
by  the  generator  action  can  also  be  determined  by  the  familiar 
three-finger  rule,  due  to  Fleming,  and  given  in  handbooks  and  ele- 
mentary books  on  electricity.  This  rule  is  useful  beause  it  empha- 
sizes the  three  mutually  perpendicular  directions,  those  of  the 
flux,  the  conductor,  and  the  relative  motion.  In  applying  this 
rule  to  a  machine  with  a  stationary  armature  one  must  remember 
that  the  direction  of  the  motion  in  Fleming's  rule  is  that  of  the 
conductor,  and  therefore  is  opposite  to  the  direction  of  the  actual 
motion  of  the  magnetic  field. 

Problem  1.  A  secondary  winding  is  placed  on  the  ring  (Fig.  1)  and  is 
connected  to  a  ballistic  galvanometer.  Let  the  number  of  turns  in  the 
secondary  winding  be  n,  the  flux  linking  with  each  turn  be  0  webers,  and 


60  THE  MAGNETIC  CIRCUIT  [ART.  25 

the  total  resistance  of  the  secondary  circuit  be  r  ohms.  Show  that  when 
the  current  in  the  primary  winding  is  reversed,  the  discharge  through 
the  galvanometer  is  equal  to  2nti>  coulombs. 

Prob.  2.  A  telephone  line  runs  parallel  to  a  direct-current  trolley 
feeder  for  20  kilometers.  When  a  current  of  100  amperes  flows  through 
the  feeder  a  flux  of  2  kilo-maxwells  threads  through  the  telephone  loop, 
per  meter  of  its  length.  What  is  the  average  voltage  induced  in  the  tele- 
phone line  when  the  current  in  the  trolley  feeder  drops  from  600  to  50  amp. 
within  0.1  sec.?  Ans.  22  volts. 

Prob.  3.  Determine  the  number  of  armature  conductors  in  series  in  a 
550  volt  homopolar  generator  of  the  axial  type,  running  at  a  peripheral 
speed  of  about  100  meters  per  sec.,  when  the  length  of  the  armature  iron 
is  50  centimeters,  and  the  flux  density  in  the  air-gap  is  between  18  and  19 
kilolines  per  sq.  cm.  Note:  For  the  construction  of  the  machine  see  the 
Standard  Handbook,  index,  under  "  Generators,  homopolar." 

Ans.    6. 

Prob.  4.  Draw  schematically  the  armature  and  the  field  windings  of  a 
shunt-wound  direct-current  generator,  select  a  direction  of  .rotation,  and 
show  how  to  connect  the  field  leads  to  the  brushes  so  that  the  machine 
will  excite  itself  in  the  proper  direction. 

Prob.  5.  From  a  given  drawing  of  a  direct-current  motor  predict  its 
direction  of  rotation. 

Prob.  6.  In  an  interpole  machine  the  average  reactance  voltage  per 
commutator  segment  during  the  reversal  of  the  current  is  calculated  to  be 
equal  to  34  volts.  What  is  the  required  net  axial  length  of  the  commu- 
tating  pole  to  compensate  for  this  voltage  if  the  peripheral  speed  of  the 
machine  is  65  meters  per  second,  and  the  flux  density  under  the  pole  is 
6  kl.  sq.cm.?  The  armature  winding  has  two  turns  per  commutator  seg- 
ment. Ans.  22  cm. 

25.  The  Induced  E.M.F.  in  a  Transformer.  The  three  types 
of  transformers  used  in  practice  are  shown  in  Figs.  12,  13,  and  14. 
Considering  the  iron  core  as  a  magnetic  link,  and  a  set  of  primary 
and  secondary  coils  as  an  electric  link,  one  may  say  that  the  core- 
type  transformer  has  one  magnetic  link  and  two  electric  links; 
the  shell-type  has  one  electric  link  and  two  magnetic  links;  the 
combination  or  cruciform  type  has  one  electric  and  four  magnetic 
links.  Still  another  type,  not  used  in  practice,  can  be  obtained 
from  the  core-type  by  adding  two  or  more  electric  links  to  the  same 
magnetic  link.  Each  electric  link  is  understood  to  consist  of  two 
windings:  the  primary  and  the  secondary. 

When  the  primary  winding  is  connected  to  a  source  of  constant- 
potential  alternating  voltage  and  the  secondary  winding  is  con- 
nected to  a  load,  alternating  currents  flow  in  both  windings  and  an 
alternating  magnetic  flux  is  established  in  thef  iron  core.  If  the 


CHAP.  IV] 


INDUCED  E.M.F. 


61 


Core     ,f 


primary  electric  circuit,  that  is,  the  one  connected  to  the  source  of 
power,  were  perfect,  that  is,  if  it  possessed  no  resistance  and  no 
reactance,  the  alternating  magnetic  flux  in  the  core  would  be  the 
same  at  all  loads.  It  would  have  such  a  magnitude  that  at  an}*- 
instant  the  counter-e.m.f.  induced  by  it  in  the  primary  wind- 
ing would  be  practically  equal  and  opposite  to  the  impressed 
voltage.  In  reality  the  resistance  and  the  leakage  reactance  of 
ordinary  commercial  transformers  are  so  low  that  for  the  purposes 
of  calculating  the  magnetic  circuit  the  primary  impedance  drop  may 
be  disregarded,  and  the  mag- 
netic flux  considered  constant 
and  independent  of  the  load. 

If  the  primary  applied  volt- 
age varies  according  to  the  sine 
law,  which  condition  is  nearly 
fulfilled  in  ordinary  cases,  the 
counter-e.m.f.,  which  is  practi- 
cally equal  and  opposite  to  it, 
also  follows  the  same  law. 
Hence,  according  to  eq.  (25), 
the  magnetic  flux  must  vary 
according  to  the  cosine  law, 
because  .  the  derivative  of  the 
cosine  is  minus  the  sine.  In 
other  words,  both  the  flux  and 
the  induced  e.m.f.  vary  accord-  FIG.  12.—  A  core-type  transformer. 
ing  to  the  sine  law,  but  the  two 

sine  waves  are  in  time  quadrature  with  each  other.  When  the 
flux  reaches  its  maximum  its  rate  of  change  is  zero,  and  therefore 
the  counter-e.m.f.  is  zero.  When  the  flux  passes  through  zero  its 
rate  of  change  with  the  timers  a  maximum,  and  therefore  the 
induced  voltage  at  this  instant  is  a  maximum. 

Let  0m  be  the  maximum  value  of  the  flux  in  the  core,  in  webers, 
and  let  /  be  the  frequency  of  the  supply  in  cycles  per  second. 
Then  the  flux  at  any  instant  t  is  $  =  $mcos  2nft,  and  the  e.m.f. 
induced  at  this  moment,  per  turn  of  the  primary  or  secondary 
winding  is 


e=  - 


sn 


Thus,  the  maximum  value  of  the  induced  voltage  per  turn  is 


62 


THE  MAGNETIC  CIRCUIT 


[ART.  25 


27z/$m;  hence,  the  effective  value  is  2;r/0TO/v2  =  4.44/0m.  Let 
there  be  NI  primary  turns  in  series;  the  total  primary  voltage  is 
then  equal  to  NI  times  the  preceding  value.  Expressing  the  flux 
in  megalines  we  therefore  obtain  the  following  practical  formula 
for  the  induced  voltage  in  a  transformer: 

10-2 (28) 


Coils 


Coils 


FIG.  13.— A  shell-type 
transformer. 


Mica 
Shields 


FIG.  14. — A  cruciform-type 
transformer. 


In  practice,  E\  is  assumed  to  be  equal  and  opposite  to  the  applied 
voltage  (for  calculating  the  flux  only,  but  not  for  determining  the 
voltage  regulation  of  the  transformer).  Formula  (28)  holds  also 
for  the  secondary  induced  voltage  E%  if  the  number  of  secondary 
turns  in  series  N2  be  substituted  for  NI.  The  voltage  per  turn  is 
the  same  in  the  primary  and  in  the  secondary  winding;  therefore, 
the  ratio  of  the  induced  voltages  is  equal  to  that  of  the  number 


CHAP.  IV]  INDUCED  E.M.F.  63 

of  turns  in  the  primary  and  secondary  windings:  that  is,  we 
have  El:E2  =  N1:N2. 

Prob.  7.  A  60-cycle  transformer  is  to  be  designed  so  as  to  have  a  flux 
density  in  the  core  of  about  9  kl./sq.cm.;  the  difference  of  potential 
between  consecutive  turns  must  not  exceed  5  volts.  What  is  the  required 
cross-section  of  the  iron?  Ans.  210  sq.cm. 

Prob.  8.  The  transformer  in  the  preceding  problem  is  to  be  wound  for 
6600  v.  primary,  and  440  v.  secondary.  What  are  the  required  numbers 
of  turns?  Ans.  1320  and  88. 

Prob.  9.  Referring  to  the  transformer  in  the  preceding  problem,  what 
are  the  required  numbers  of  turns  if  three  such  transformers  are  to  be  used 
Y-connected  on  a  three-phase  system,  for  which  the  line  voltages  are  6600 
and  440  respectively?  Ans.  765  and  51. 

Prob.  10.  In  a  110-kilovolt,  25-cycle  transformer  for  Y-connection  the 
net  cross-section  of  the  iron  is  about  820  sq.cm.  and  the  permissible  maxi- 
mum flux  density  is  about  10.7  kl/sq.cm.  What  is  the  number  of  turns 
in  the  high-tension  winding?  Ans.  6500. 

Prob.  11.  The  secondary  of  the  transformer  in  the  preceding  problem 
is  to  be  wound  for  6600  v.,  delta  connection,  with  taps  for  varying  the 
secondary  voltage  within  i5  per  cent.  Specify  the  winding. 

Ans.    709  turns;  taps  taken  after  the  34th  and  68th  turn. 

Prob.  12.  Explain  the  reason  for  which  a  60-cycle  transformer  usually 
runs  hot  even  at  no  load,  when  connected  to  a  25-cycle  circuit  of  the  same 
voltage.  Show  from  the  core-loss  curves  that  the  voltage  must  be 
reduced  to  from  75  to  85  per  cent  of  its  rated  value  in  order  to  have  the 
normal  temperature  rise  in  the  transformer,  at  the  rated  current. 

Prob.  13.  Show  graphically  that  the  wave  of  the  flux,  within  a  trans- 
former, becomes  more  and  more  peaked  when  the  wave  of  the  applied  e.m.f. 
becomes  more  and  more  flat,  and  vice  versa.  Hint :  The  instantaneous 
values  of  e.m.f.  are  proportional  to  the  values  of  the  slope  of  the  curve  of 
flux. 

Prob.  14.  The  wave  of  the  voltage  impressed  upon  a  transformer  has 
a  15  per  cent  third  harmonic  which  flattens  the  wave  symmetrically.  Show 
analytically  that  the  corresponding  flux  wave  has  a  5  per  cent  third  har- 
monic in  such  a  phase  position  as  to  make  the  flux  wave  peaked. 

26.  The  Induced  E.M.F.  in  an  Alternator  and  in  an  Induction 
Motor.  Part  of  a  revolving  field  alternator  is  shown  in  Fig.  15. 
The  armature  core  is  stationary  and  has  a  winding  placed  in  slots, 
which  may  be  either  open  or  half  closed.  The  pole  pieces  are 
mounted  on  a  spider  and  are  provided  with  an  exciting  winding. 
When  the  spider  is  driven  by  a  pime-mover  the  magnetic  flux 
sweeps  past  the  armature  conductors  and  induces  alternating 
voltages  in  them.1  In  order  to  obtain  an  e.m.f.  approaching  a  sine 

1  For  details  concerning  the  different  types  of  armature  windings  see  the 
author's  Experimental  Electrical  Engineering,  Vol.  2,  Chap.  30. 


64 


THE  MAGNETIC  CIRCUIT 


[ART.  26 


wave  as  nearly  as  possible  the  pole  shoes  are  shaped  as  shown 
in  the  sketch,  that  is  to  say,  so  as  to  make  a  variable  air-gap  and 
thus  grade  the  flux  density  from  the  center  of  the  pole  to  the  edges. 
In  high-speed  turbo-alternators  the  field  structure  often  has  a 
smooth  surface,  without  projecting  poles  (Fig.  33),  in  order  to 
reduce  the  noise  and  the  windage  loss.  Such  a  structure  is  also 
stronger  mechanically  than  one  with  projecting  poles.  The  grad- 
ing of  the  flux  is  secured  by  distributing  the  field  winding  in  slots, 
so  that  the  whole  m.m.f .  acts  on  only  part  of  the  pole  pitch. 

Consider  a  conductor  at  a  during  the  interval  of  time  during 
which  the  flux  moves  by  one  pole  pitch  r.     The  average  e.m.f. 


FIG.  15. — The  cross-section  of  a  synchronous  machine. 

induced  in  the  conductor  is,  according  to  eq.  (26),  equal  to 
where  4>  is  the  total  flux  per  pole  in  webers,  and  T  is  the  time  of 
one  complete  cycle,  corresponding  to  2r  the  space  of  two  pole 
pitches.  But  T=l/f,  so  that  the  average  voltage  induced  in  a 
conductor  is 


,=  2/0. 


(29) 


The  value  of  eave  thus  does  not  depend  upon  the  distribution  of  the 
flux  0  in  the  air-gap. 

If  the  pole-pieces  are  shaped  so  as  to  give  an  approximately 
sinusoidal  distribution  of  flux  in  the  air-gap,  the  induced  e.m.f.  is 
also  approximately  a  sine  wave,  and  the  ratio  between  the  effect- 


CHAP.  IV]  INDUCED  E.M.F.  65 


ive  and  the  average  values  of  the  voltages  is  equal  to 
or  1.1  1.1  If  the  shape  of  the  induced  e.m.f.  departs  widely  from  the 
sine  wave  the  actual  curve  must  be  plotted  and  its  form  factor 
determined  by  one  of  the  known  methods  (see  the  Electric  Cir- 
cuit). Let  the  form  factor  in  general  be  7  and  let  the  machine 
have  N  armature  turns  in  series  per  phase,  or  what  is  the  same, 
2N  conductors  in  series.  The  total  induced  e.m.f.  in  effective 
volts  is  then 

(30) 


This  formula  presupposes  that  there  is  but  one  slot  per  pole  per 
phase,  so  that  the  e.m.fs.  induced  in  the  separate  conductors  are  all 
in  phase  with  each  other,  and  that  their  values  are  simply  added 
together.  In  reality,  there  is  usually  more  than  one  slot  per  pole 
per  phase,  for  practical  reasons  discussed  in  the  next  article.  It 
will  be  seen  from  the  figure  that  the  e.m.fs.  induced  in  adjacent 
slots  are  somewhat  out  of  phase  with  each  other,  because  the  crest 
of  the  flux  reaches  different  slots  at  different  times.  Therefore, 
the  resultant  voltage  of  the  machine  is  somewhat  smaller  than  that 
according  to  the  preceding  formula.  The  influence  of  the  dis- 
tribution of  the  winding  in  the  slots  is  taken  into  account  by  mul- 
tiplying the  value  of  E  in  the  preceding  formula  by  a  coefficient  kb, 
which  is  smaller  than  unity  and  which  is  called  the  breadth  factor. 
Introducing  this  factor,  and  assuming  7  =  1.11,  which  is  accurate 
enough  for  good  commercial  alternators,  we  obtain 


(31) 


where  $  is  now  in  megalines.     Values  of  kb  are  given  in  the  articles 
that  follow. 

Formula  (31)  applies  equally  well  to  the  polyphase  induction 
motor  or  generator.  There  we  also  have  a  uniformly  revolving 
flux  in  the  air-gap,  the  flux  density  being  distributed  in  space, 
according  to  the  sine  law.  This  gliding  flux  induces  e.m.fs.  in  the 
stator  and  rotor  windings.  The  only  difference  between  the  two 
kinds  of  machines  is  that  in  the  synchronous  alternator  the  field  is 
made  to  revolve  by  mechanical  means,  while  in  an  induction 
machine  the  field  is  excited  by  the  polyphase  currents  flowing  in 

1  For  the  proportions  of  a  pole-shoe  which  very  nearly  give  a  sine  wave 
see  Arnold,  Wechselstromtechnik,  Vol.  3,  p.  247. 


66  THE  MAGNETIC  CIRCUIT  [ART.  27 

the  stator  and  rotor  windings.  The  formulae  of  this  chapter  also 
apply  without  change  to  the  synchronous  motor,  because  the  con- 
struction and  the  operation  of  the  latter  are  identical  with  those  of 
an  alternator;  the  only  difference  being  that  an  alternator  trans- 
forms mechanical  energy  into  electrical  energy,  while  a  synchro- 
nous motor  transforms  energy  in  the  reverse  direction.  In  all 
cases  the  induced  voltage  is  understood  and  not  the  line  voltage. 
The  latter  may  differ  considerably  from  the  former,  due  to  the 
impedance  drop  in  the  stator  winding. 

Prob.  15.  A  delta-connected,  2300-v.,  60-cycle,  128.5-r.p.m.  alternator 
is  estimated  to  have  a  useful  flux  of  about  3.9  megalines  per  pole.  If  the 
machine  has  one  slot  per  pole  per  phase  how  many  conductors  per  slot  are 
needed?  Ans.  8. 

Prob.  16.  A  100,000-cycle  alternator  for  wireless  work  has  one  conduc- 
tor per  pole  and  600  poles.  The  rated  voltage  at  no  load  is  110  v.  What 
is  the  flux  per  pole  and  the  speed  of  the  machine? 

Ans.    82.5  maxwells;  20,000  r.p.m. 

Prob.  17.  It  is  desired  to  design  a  line  of  induction  motors  for  a  per- 
ipheral speed  of  50  met.  per  sec.,  the  maximum  density  in  the  air-gap  to  be 
about  6  kilolines  per  sq.cm.  What  will  be  the  maximum  voltage  induced 
per  meter  of  active  length  of  the  stator  conductors?  Hint:  Use  formula 
(27).  Ans.  30  volt. 

Prob.  18.  Formula  (31)  is  deduced  under  the  assumption  that  each 
armature  conductor  is  subjected  to  the  "  cutting  "  action  of  the  whole 
flux.  In  reality,  almost  the  whole  flux  passes  through  the  teeth  between 
the  conductors,  so  that  it  may  seem  upon  a  superficial  inspection  that 
little  voltage  could  be  induced  in  the  conductors  which  are  embedded  in 
slots.  Show  that  such  is  not  the  case,  and  that  the  same  average  voltage 
is  induced  in  the  conductors  placed  in  completely  closed  slots,  as  in  the 
conductors  placed  on  the  surface  of  a  smooth-body  armature.  Hint: 
When  the  flux  moves,  the  same  amount  of  magnetic  disturbance  must 
pass  in  the  tangential  direction  through  the  slots  as  through  the  teeth. 

Prob.  19.  Deduce  eq.  (31)  directly  from  eq.  (27).  Can  eq  (31)  be 
derived  under  the  case  of  the  transformer  action? 

27.  The  Breadth  Factor.  Armature  conductors  are  usually 
placed  in  more  than  one  slot  per  pole  per  phase,  for  the  following 
reasons : 

(a)  The  distribution  of  the  magnetic  field  is  more  uniform, 
there  being  less  bunching  of  the  flux  under  the  teeth ; 

(b)  The  induced  e.m.f.  has  a  better  wave  form; 

(c)  The  leakage  reactance  of  the  winding  is  reduced; 

(d)  The  same  armature  punching  can  be  used  for  machines  with 
different  numbers  of  poles  and  phases; 


CHAP.  IV]  INDUCED  E.M.F.  67 

(e)  The  mechanical  arrangement  and  cooling  of  the  coils  is 
somewhat  simplified. 

The  disadvantage  of  a  large  number  of  slots  is  that  more  space 
is  taken  up  by  insulation,  and  the  machine  becomes  more  expen- 
sive, especially  if  it  is  wound  for  a  high  voltage.  The  electromo- 
tive force  is  also  somewhat  reduced  because  the  voltages  induced 
in  different  slots  are  somewhat  out  of  phase  with  one  another. 
The  advantages  of  a  distributed  winding  generally  outweigh  its  dis- 
advantages, and  such  windings  are  used  almost  entirely.  Thus, 
it  is  of  importance  to  know  how  to  calculate  the  value  of  the 
breadth  factor  kb  for  a  given  winding. 


|  Phase  1 
I  »  2 
D  »  3 

FIG.  16. — A  fractional-pitch  winding. 

In  the  winding  shown  in  Fig.  15  each  conductor  is  connected 
with  another  conductor  situated  at  a  distance  exactly  equal  to  the 
pole  pitch.  It  is  possible,  however,  to  connect  one  armature  con- 
ductor to  another  at  a  distance  somewhat  smaller  than  the  pole 
pitch  (Fig.  16).  Such  a  winding  is  called  &  fractional-pitch  wind- 
ing, in  distinction  to  the  winding  shown  in  Fig.  15;  the  latter 
winding  is  called  a  full-pitch  or  hundred-per  cent  pitch  winding. 
It  will  be  seen  from  Fig.  16  that,  with  a  two-layer  fractional-pitch 
winding,  some  slots  are  occupied  by  coils  belonging  to  two  different 
phases.  The  advantages  of  the  fractional-pitch  winding  are : 

(a)  The  end-connections  of  the  winding  are  shortened,  so  that 
there  is  some  saving  in  armature  copper. 

(6)  The  end-connections  occupy  less  space  in  the  axial  direc- 
tion of  the  machine,  so  that  the  whole  machine  is  shorter. 


68  THE  MAGNETIC  CIRCUIT  [ART.  28 

(c)  In  a  two-pole  or  four-pole  machine  it  is  necessary  to  use  a 
fractional-pitch  winding  in  order  to  be  able  to  place  machine- 
wound  coils  into  the  slots. 

A  disadvantage  of  the  fractional-pitch  winding  is  that  the 
e.m.fs.  induced  on  both  sides  of  the  same  coils  are  not  exactly  in 
phase  with  each  other,  so  that  for  a  given  voltage  a  larger  number 
of  turns  or  a  larger  flux  is  required  than  with  a  full-pitch  winding. 
Fractional-pitch  windings  are  used  to  a  considerable  extent  both 
in  direct-  and  in  alternating-current  machinery. 

Thus,  the  induced  e.m.f .  in  an  alternator  or  an  induction  motor 
is  reduced  by  the  distribution  of  the  winding  in  more  than  one 
slot,  and  also  by  the  use  of  a  fractional-winding  pitch.  It  is 
therefore  convenient  to  consider  the  breadth  factor  A;&  as  being 
equal  to  the  product  of  two  factors,  one  taking  into  account  the 
number  of  slots,  and  the  other  the  influence  of  the  winding  pitch. 
We  thus  put 

kb=kskw, (32) 

where  ks  is  called  the  slot  factor  and  kw  the  winding-pitch  factor. 

For  a  full-pitch  winding  kw=l,  and  kb=k8')  for  a  fractional- 
pitch  unislot  winding  ks  =  1,  and  Afc  =  kw.  The  factors  ks  and  kw 
are  independent  of  one  another,  and  their  values  are  calculated 
in  the  next  two  articles. 

28.  The  Slot  Factor  ks.  Let  the  stator  of  an  alternator  (or 
induction  motor)  have  two  slots  per  pole  per  phase,  and  let  the 
centers  of  the  adjacent  slots  be  displaced  by  an  angle  a,  in  electri- 
cal degrees,  the  pole  pitch,  T,  corresponding  to  180  electrical 
degrees.  If  E  (Fig.  17)  is  the  vector  of  the  effective  voltage 
induced  in  the  conductors  in  one  slot,  the  voltage  Er  due  to  the 
conductors  in  both  slots  is  represented  graphically  as  the  geometric 
sum  of  two  vectors  E  relatively  displaced  by  the  angle  a.  We  see 
from  the  figure  that  %E'  =  Ecos  %a,  or  E'  =  2E  cos  i«.  If  both  sets 
of  conductors  were  bunched  in  the  same  slot  we  would  then  have 
E'=  2E.  Hence,  in  this  case  the  coefficient  of  reduction  in 
voltage,  or  the  slot  factor,  fcs=cosja. 

Let  now  the  armature  stamping  have  S  slots  per  pole  per  phase, 
the  angle  between  adjacent  slots  being  again  equal  to  a  electrical 
degrees.  Let  the  vectors  marked  E  in  Fig.  18  be  the  voltages 
induced  in  each  slot;  the  resultant  voltage  E'  is  found  as  the  geo- 


CHAP.  IV] 


INDUCED  E.M.F. 


69 


metric  sum  of  the  E's.     The  radius  of  the  circle  r  =  %E/sm%a,  and 
\E'  =  r  sm^Sa .     Therefore, 


E'/SE=(sm 


sn 


(33) 


When  S=2,  the  formula  (33)  becomes  identical  with  the  expres- 
sion given  before. 


FIG.  17.  —  A  diagram  illustrating  the  slot    FIG.   18.  —  A,  diagram  illustrating 
factor  with  two  slots.  the  slot  factor  with  several  slots. 

The  angle  a  depends  upon  the  number  of  slots  and  the  number 
of  phases.     Let  there  be  ra  phases;  then  aSm=  180  degrees,  and 

.        .     .'    '.     .     .     .     (34) 


The  values  of  ks  in  the  table  below  are  calculated  by  using  the  for- 
mulae (33)  and  (34). 

VALUES   OF  THE  SLOT  FACTOR  ks 


Slots  per  Phase 
per  Pole. 

Single-phase 
Winding. 

Two-phase 
Winding. 

Three-phase 
Winding. 

1 

1.000 

1.000 

1.000 

2 

0.707 

0.924 

0.966 

3 

0.667 

0.911 

0.960 

4 

0.653 

0.907- 

0.958 

5 

0.647 

0.904 

0.957 

6 

0.643 

0.903 

0.956 

Infinity 

0.637 

0.900 

0.955 

In  single-phase  alternators  part  of  the  slots  are  often  left  empty 
so  as  to  reduce  the  breadth  of  the  winding  and  therefore  increase 
the  value  of  ks.  For  instance,  if  a  punching  is  used  with  six  slots 
per  pole,  perhaps  only  three  or  four  adjacent  slots  are  occupied. 
In  this  case,  it  would  be  wrong  to  take  the  values  of  ks  from  the 
first  column  of  the  table.  If,  for  instance,  three  slots  out  of  six 


70  THE  MAGNETIC  CIRCUIT  [ABT.  29 

are  occupied,  the  value  of  ks  is  the  same  as  for  a  two-phase  wind- 
ing with  three  slots  per  pole  per  phase. 

Prob.  20.  Check  some  of  the  values  of  ks  given  in  the  table  above. 

Prob.  21.  The  armature  core  of  a  single-phase  alternator  is  built  of 
stampings  having  three  slots  per  pole;  two  slots  per  pole  are  utilized. 
What  is  the  value  of  kst  Ans.  0.866. 

Prob.  22.  A  single-phase  machine  has  S  uniformly  distributed  slots  per 
pole,  of  which  only  S'  are  used  for  the  winding.  What  is  the  value  of  /bs? 
Ans.  Use  S'  in  eq.  (33)  instead  of  S',  preserve  S  in  eq.  (34). 

Prob.  23.  A  six-pole,  6600  v.,  Y-connected,  50-cycle  turbo-alternator  is 
to  be  built,  using  an  armature  with  90  slots.  The  estimated  flux  per  pole 
is  about  6  megalines.  How  many  conductors  are  required  per  slot? 

Ans.  20. 

Prob.  24.  What  is  the  value  of  ks  when  the  winding  is  distributed  uni- 
formly on  the  surface  of  a  smooth-body  armature,  each  phase  covering  /? 
electrical  degrees?  Solution :  Referring  to  Fig.  18,  ks  is  in  this  case  equal 
to  the  ratio  of  the  chord  E'  to  the  arc  of  the  circle  which  it  subtends.  The 
central  angle  is  /?,  and'  we  have  ks  =  (sin^/?)  /  (%px/ 180°) .  In  a  three-phase 
machine  /?  =  60  degrees,  and  therefore  ks  =  0.955.  This  is  the  value  given 
in  the  last  column  of  the  table  above. 

Prob.  25.  Deduce  the  expression  for  ks  given  in  the  preceding  problem 
directly  from  formula  (33).  Solution:  Substituting  $«  =  /?;  S  =  00 and 
a  =0,  an  indeterminate  expression,  O.oo,  is  obtained.  But  when  the 
angle  a  approaches  zero  its  sine  is  nearly  equal  to  the  arc,  so  that  the 
denominator  of  the  right-hand  side  of  eq.  (33)  approaches  the  value 
iS.ia  =  i/?,  where  /?  is  in  radians.  Changing  /?  to  degrees,  the  required 
formula  is  obtained. 

29.  The  Winding-pitch  Factor  kw.  Let  the  distance  between 
the  two  opposite  sides  of  a  coil  (Fig.  16)  be  180  —7-  degrees,  where 
Y  is  the  angle  by  which  tfye  winding-pitch  is  shortened.  The  volt- 
ages induced  in  the  two  sides  of  the  coil  are  out  of  phase  with 
each  other  by  the  angle  7-,  so  that  if  the  voltage  induced  in  each 
side  is  e,  the  total  voltage  is  equal  to  2e  Cosjf  (Fig.  17).  Fig.  17. 
will  apply  to  this  case  if  we  read  f  for  the  angle  a.  Hence,  we 
have  that 

kw=  cos  \Y (35) 

In  practice,  the  winding-pitch  is  measured  in  per  cent,  or  as  a  frac- 
tion of  the  pole  pitch  r.  For  instance,  if  there  are  nine  slots  per 
pole  and  the  coil  lies  in  slots  1  and  8,  the  winding-pitch  is  7/9,  or 
77.8  per  cent.  If  the  coil  were  placed  in  slots  1  and  10  we  would 
have  a  full-pitch,  or  a  100  per  cent  pitch  winding.  Let  in  gen- 


CHAP.  IV] 


INDUCED  E.M.F. 


71 


eral  the  winding-pitch  be  £,  expressed  as  a  fraction.  Then 
Y=  (1  —  Q180°.  Substituting  this  value  of  f  into  formula  (35)  we 
obtain 


/cu,=cos[900(l-Q] 


(36) 


The  values  of  kw  given  in  Fig.  19  have  been  calculated  according 
to  this  formula. 


0.90 


0.80 


0.70 


50 


70 


90 


100 


Per  Cent  Winding  Pitch 
FIG.  19. — Values  of  the  winding-pitch  factor. 

In  applications,  one  takes  the  value  of  k8  from  the  table,  assum- 
ing the  winding  pitch  to  be  one  hundred  per  cent,  and  multiplies  it 
by  the  value  of  kw  taken  from  the  curve  (Fig.  19).  With  frac- 
tional-pitch two-layer  windings  the  value  of  ks  corresponds  to  the 
number  of  slots  per  layer  per  pole  per  phase,  and  not  to  the  total 
number  of  slots  per  pole  per  phase.  This  is  clear  from  the  explana- 
tion given  in  the  preceding  paragraph.  Thus/for  instance,  in  Fig. 
16,  k8  must  be  taken  for  three  slots  and  not  for  five  slots.  If  one 
has  to  calculate  the  values  of  kb  often,  it  is  advisable  to  plot  a  set 


72  THE   MAGNETIC  CIRCUIT  [ART.  30 

of  curves,  like  the  one  in  Fig.  19,  each  curve  giving  the  values  of  kb 
for  a  certain  number  of  slots  per  pole  per  phase,  against  per  cent 
winding  pitch  as  abscissae. 

Prob.  26.  In  a  4-pole,  72-slot,  turbo-alternator  the  coils  lie  in  slots  1 
and  13.  What  is  the  per  cent  winding-pitch  and  by  what  percentage  is 
the  e.m.f.  reduced  by  making  the  pitch  short  instead  of  100%? 

Ans.  66.7  per  cent;  l—kw  =  13.4  per  cent. 

Prob.  27.  What  is  the  flux  per  pole  at  no  load  in  a  6600-volt,  25-cycle, 
500-r.p.m.,  Y-connected  induction  motor  which  has  90  slots,  36  conduc- 
tors per  slot,  and  a  winding-pitch  of  about  73  per  cent? 

Ans.  7.26  megalines. 

Prob.  28.  Show  that  for  a  chain  winding  kw  is  always  equal  to  unity, 
in  spite  of  the  fact  that  some  of  the  coils  are  narrower  than  the  pole  pitch. 

Prob.  29.  Draw  a  sketch  of  a  single-layer,  fractional-pitch  winding, 
using  alternate  slots  for  the  overlapping  phases.  Show  what  values  of  ka 
and  kw  should  be  used  for  such  a  winding. 

30.  Non-sinusoidal  Voltages.  In  the  foregoing  calculations 
the  supposition  is  made  that  the  flux  density  in  the  air-gap  is  dis- 
tributed according  to  the  sine  law  so  that  sinusoidal  voltages  are 
induced  in  each  conductor.  Under  these  circumstances  the 
resultant  voltage  also  follows  the  sine  law,  no  matter  what  the 
winding-pitch  and  the  number  of  slots  are.  The  flux  is  practically 
sinusoidal  in  induction  motors  because  the  higher  harmonics  of 
the  flux  are  wiped  out  by  the  secondary  currents  induced  in  the 
low-resistance  rotor.  But  in  synchronous  alternators  and  motors 
with  projecting  poles  the  distribution  of  the  flux  in  the  air-gap  is 
usually  different  from  a  pure  sine  wave.  For  instance,  when  the 
pole  shoe  is  shaped  by  a  cylindrical  surface  concentric  with  that  of 
the  armature,  the  air-gap  length  and  consequently  the  flux  density 
are  constant  over  the  larger  portion  of  the  pole;  therefore,  the 
curve  of  the  field  distribution  is  a  flat  one.  This  shape  is  improved 
to  some  extent  by  chamfering  the  pole-tips  or  by  shaping  the  pole 
shoes  to  a  circle  of  a  smaller  radius,  so  that  the  length  of  the  air-gap 
increases  gradually  toward  the  pole-tips. 

When  a  machine  revolves  at  a  uniform  speed,  the  e.m.f.  induced 
in  a  single  armature  conductor  has  exactly  the  shape  of  the  field- 
distribution  curve,  because  in  this  case  the  rate  of  cutting  the  flux 
is  proportional  to  the  flux  density  (see  eq.  27  above).  There- 
fore, when  a  machine  has  but  one  slot  per  pole  per  phase  (which 
condition  is  undesirable,  but  unavoidable  in  low-speed  alternators, 
or  in  those  designed  for  extremely  high  frequencies),  the  shape  of 


CHAP.  IV]  INDUCED  E.M.F.  73 

the  pole-pieces  must  be  worked  out  very  carefully  in  order  to  have 
an  e.m.f.  approaching  the  true  sine  wave.  With  a  larger  number 
of  slots  this  is  not  so  necessary  because  the.  em.fs.  induced  in  differ- 
ent slots  are  added  out  of  phase  with  each  other,  and  the  undesir- 
able higher  harmonics  partly  cancel  each  other.  The  voltage 
wave  is  further  improved  by  a  judicious  use  of  a  fractional-pitch 
winding.  These  facts  are  made  clearer  in  the  solution  of  the  prob- 
lems that  follow.1 

Prob.  30.  The  flux  density  in  the  air-gap  under  the  poles  of  an  alterna- 
tor is  constant  for  50  per  cent  of  the  pole  pitch,  and  then  it  drops  to  zero, 
according  to  the  straight-line  law,  on  each  side  in  a  space  of  15  per  cent 
of  the  pole  pitch.  Draw  to  scale  the  curves  of  induced  e.m.f.  for  the  fol- 
lowing windings :  (a)  Single-phase,  one  slot  per  pole ;  (b)  Single-phase, 
nine  slots  per  pole,  five  slots  being  occupied  by  a  one-hundred  per  cent 
pitch  winding ;  (c)  The  same  as  in  (b)  only  the  winding-pitch  is  equal  to 
7/9;  (d)  Three-phase,  Y-connected  full-pitch  winding,  two  slots  per  pole 
per  phase;  in  the  latter  case  give  curves  of  both  the  phase  voltage  and  the 
line  voltage.  On  all  the  curves  indicate  roughly  the  equivalent  sine  wave, 
in  order  to  see  the  influence  of  the  number  of  slots  and  of  the  fractional 
pitch  in  improving  the  wave  form. 

Prob.  31.  A  three-phase,  Y-connected  alternator  has  three  slots  per 
pole  per  phase,  and  a  full-pitch  winding.  The  field  curve  has  an  8  per 
cent  fifth  harmonic,  that  is  to  say,  the  amplitude  of  the  fifth  harmonic  is 
0.08  of  that  of  the  fundamental  sine  wave.  What  is  the  magnitude  of  the 
fifth  harmonic  in  the  phase  voltage  and  in  the  line  voltage.  Solution :  In 
formula  (33)  the  angle  a  between  the  adjacent  slots  is  20  electrical  degrees 
for  the  fundamental  wave.  For  the  fifth  harmonic  the  same  distance 
between  the  slots  corresponds  to  100  electrical  degrees.  Hence,  for  the 
fundamental  wave 

fcs  =  sin  30°/(3  sin  10°)  =0.96; 

while  for  the  fifth  harmonic 

ks6  =  sm  150°/(3  sin  50°)  =0.217. 

This  means  that,  due  to  the  distribution  in  three  slots,  the  fundamental 
wave  of  the  voltage  is  reduced  to  0.96  of  its  value  in  a  unislot  machine, 
while  the  fifth  harmonic  is  reduced  to  only  0.217  of  its  corresponding  value. 
Therefore,  the  relative  magnitude  of  the  fifth  harmonic  in  the  phase 
voltage  is  8  X  21. 7/96  =  1.8  per  cent,  which  means  that  the  fifth  harmonic 
is  reduced  to  less  than  one-fourth  of  its  value  in  the  field  curve.  In 
calculating  the  line  voltage  the  vectors  of  the  fundamental  waves  in  a 
three-phase  machine  are  combined  at  an  angle  of  120  degrees.  Conse- 

1  For  further  details  see  Professor  C.  A.  Adams'  paper  on  "  Electromotive 
Force  Wave-shape  in  Alternators,"  Trans.  Amer.  Inst.  Elec.  Engs.,  Vol. 
28  (1909),  Part  II,  p.  1053. 


74  THE  MAGNETIC  CIRCUIT  [ART.  31 

quently,  the  vectors  of  the  fifth  harmonic  are  combined  at  an  angle 
of  120X5=600  degrees,  or  what  is  the  same,  — 120  degrees.  Therefore 
the  proportion  of  the  fifth  harmonic  in  the  line  voltage  is  the  same  as 
that  in  the  phase  voltage. 

Prob.  32.  Solve  the  foregoing  problem  when  the  winding  pitch  is  7/9. 
Ans.  0.33  per  cent.  This  shows  that  by  properly  selecting  the 
winding  pitch  an  objectionable  higher  harmonic  can  be 
reduced  to  a  negligible  amount. 

Prob.  33.  Show  that  the  line  voltage  of  a  Y-connected  machine  can 
have  no  3d,  9th,  15th,  etc.  harmonics,  that  is  to  say,  harmonics  the  num- 
bers of  which  are  multiples  of  3,  no  matter  to  what  extent  such  harmonics 
are  present  in  the  induced  e.m.fs.  in  each  phase. 

Prob.  34.  Prove  that  in  order  to  have  even  harmonics  in  the  induced 
e.m.f .  of  an  alternator  two  conditions  are  necessary :  (a)  the  flux  distribu- 
tion under  the  alternate  poles  must  be  different;  (b)  the  distribution  of 
the  armature  conductors  under  the  alternate  poles  must  also  be  different 
from  one  another.  Indicate  pole  shapes  and  an  arrangement  of  the  arma- 
ture winding  particularly  favorable  for  the  production  of  the  second  har- 
monic. Note:  In  spite  of  a  different  distribution  of  flux  densities  the 
total  flux  is  the  same  under  all  the  poles.  Therefore,  the  average  voltages 
for  both  half  cycles  are  equal  (see  Art.  24),  though  the  shape  of  the  two 
halves  of  the  curve  may  be  different,  due  to  the  presence  of  even  har- 
monics. This  shows  that  there  is  no  "  continuous-voltage  component  " 
in  the  wave,  or  rather  that  the  voltage  is  in  no  sense  unidirectional,  and 
that  a  direct-current  machine  cannot  be  built  with  alternate  poles  without 
the  use  of  some  kind  of  a  commutating  device. 

31.  The  Induced  E.M.F.  in  a  Direct-current  Machine.     The 

e.m.f.  induced  in  the  armature  coils  of  a  direct-current  machine 
(Fig.  20)  is  alternating,  but  due  to  the  commutator,  the  voltage 
between  the  brushes  of  opposite  polarity  remains  constant. 
This  voltage  is  equal  at  any  instant  to  the  sum  of  the  instantaneous 
e.m.fs.  induced  in  the  coils  which  are  connected  in  series  between 
the  brushes.  When  a  coil  is  transferred  from  one  circuit  to 
another,  a  new  coil  in  the  same  electromagnetic  position  is  intro- 
duced into  the  first  circuit,  and  in  this  wise  the  voltage  between 
the  brushes  is  maintained  practically  constant,  except  for  the  small 
variations  which  occur  while  the  armature  is  coming  back  to  a 
symmetrical  position.  These  variations  are  due  to  the  coils  short- 
circuited  by  the  brushes  and  to  the  fact  that  the  number  of 
commutator  segments  is  finite. 

Thus,  to  obtain  the  value  of  the  voltage  between  the  brushes, 
it  is  necessary  to  find  the  sum  of  the  e.m.fs.  induced  at  some 
instant  in  the  individual  armature  coils  which  are  connected  in 


CHAP.  IV] 


INDUCED  E.M.F. 


75 


series  between  the  brushes.  Each  e.m.f.  represents  an  instanta- 
neous value  of  an  alternating  e.m.f.;  the  e.m.fs.  induced  in  different 
coils  differing  in  phase  from  one  another,  because  they  occupy 
different  positions  with  respect  to  the  poles.  The  voltages  induced 
in  the  extreme  coils  of  an  armature  circuit  differ  from  one  another 
by  one-half  of  a  cycle. 

Instead  of  adding  the  actual  instantaneous  voltages,  it  is  suffi- 
cient to  calculate  the  average  voltage  per  coil,  and  to  multiply  it 
by  the  number  of  coils  in  series,  because  the  wave  form  of  the 
e.m.fs.  induced  in  all  the  coils  is  the  same,  and  their  phase  differ- 


FIG.  20. — The  cross-section  of  a  direct-current  machine. 

ence  is  distributed  uniformly  over  one-half  of  a  cycle.  According 
to  eq.  (26)  the  average  voltage  per  turn  per  half  a  cycle  is  24>/%T, 
where  %T  is  the  time  during  which  the  coil  moves  by  one  pole 
pitch,  and  $  is  the  flux  per  pole,  in  webers.  Substituting  I//  for 
T,  the  average  voltage  per  turn  is  equal  to  4/0.  Let  there  be  N 
turns  in  series  between  the  brushes  of  opposite  polarity;  then  the 
induced  voltage  of  the  machine  is 


1-2 


(37) 


where  0  is  now  in  megalines.  Thus,  in  a  direct-current  machine 
the  induced  voltage  between  the  brushes  depends  only  upon  the 
total  useful  flux  per  pole,  and  not  upon  its  distribution  in  the  air- 
gap. 


76  THE  MAGNETIC  CIRCUIT  [ART.  31 

The  relation  between  the  number  of  turns  in  series  and  the  total 
number  of  turns  on  the  armature  depends  upon  the  kind  of  the 
armature  winding.1  If  the  armature  has  a  multiple  winding  N  is 
equal  to  the  total  number  of  turns  on  the  armature  divided  by  the 
number  of  poles.  For  a  two-circuit  winding  the  number  of  turns 
in  series  is  equal  to  one-half  of  the  total  number  of  turns.  The  num- 
ber of  poles  and  the  speed  of  the  machine  do  not  enter  explicitly 
into  formula  (37),  but  are  contained  in  the  value  of  /. 

Prob.  35.  A  90-slot  armature  is  to  be  used  for  a  6-pole,  580-r.p.m.,  250- 
v.,  direct-current  machine  with  a  multiple  winding.  How  many  conduc- 
tors per  slot  are  necessary  if  the  permissible  flux  per  pole  is  about  3  mega- 
lines?  Ans.  10. 

Prob.  36.  A  550-v.,  4-pole  railway  motor  has  a  two-circuit  armature 
winding  which  consists  of  59  coils,  8  turns  per  coil.  The  total  resistance 
of  the  motor  is  0.235  ohm.  When  the  motor  runs  at  675  r.p.m.  it  takes  in 
81  amp.  What  is  the  flux  per  pole  at  this  load?  Aris.  2.5  ml. 

Prob.  37.  Show  that  in  a  direct-current  machine  the  use  of  a  fractional- 
pitch  winding  has  no  effect  whatever  upon  the  value  of  the  induced  e.m.f  ., 
as  long  as  the  winding-pitch  somewhat  exceeds  the  width  of  the  pole  shoe. 

Prob.  38.  Prove  that  formula  (37)  is  identical  with  the  expression 


(38) 


where  C  is  the  total  number  of  armature  conductors,  p  is  the  number  of 
poles,  and  p'  is  the  number  of  circuits  in  parallel. 

Prob.  39.  Show  that  the  induced  e.m.f.  is  the  same  when  the  armature 
conductors  are  placed  in  open  or  in  closed  slots  as  when  they  are  on  the 
surface  of  a  smooth-body  armature.  See  Prob.  18,  Art.  26. 

Prob.  40.  Considerable  effort  has  been  made  to  produce  a  direct-cur- 
rent generator  with  alternate  poles,  and  without  any  commutator.  One  of 
the  proposals  which  is  sometimes  urged  by  a  beginner  is  to  use  an  ordinary 
alternator,  and  to  supply  the  exciting  winding  with  an  alternating  current 
of  the  synchronous  frequency.  The  apparent  reasoning  is  that  the  field 
being  reversed  at  the  completion  of  one  alternation  the  next  half  wave  of 
the  induced  e.m.f.  must  be  in  the  same  direction  as  the  preceding  one,  thus 
giving  a  unidirectional  voltage.  Show  that  such  a  machine  in  reality 
would  give  an  ordinary  alternating  voltage  of  double  the  frequency. 
Hint  :  Make  use  of  the  fact  that  an  alternating  field  can  be  replaced  by  two 
constant  fields  revolving  in  opposite  directions.  Or  else  give  a  rigid 
mathematical  proof  by  considering  the  actual  rate  at  which  the  armature 
conductors  are  cut  by  the  field,  which  field  is  at  the  same  time  pulsating 
and  revolving. 

1  For  details  concerning  the  direct-current  armature  windings  see  the 
author's  Experimental  Electrical  Engineering,  Vol.  2,  Chapter  30. 


CHAP.  IV.]  INDUCED  E.M.F.  77 

Prob.  41.  Prove  that  if  in  the  preceding  problem  the  frequency  of  the 
rotation  of  the  poles  is  flt  and  the  frequency  of  the  alternating  current  in 
the  exciting  winding  is/2,  that  the  voltage  induced  in  the  armature  is  a  com- 
bination of  two  waves  having  frequencies  of  fi  +/2  and  fi  — /2  respectively. 

32.  The  Ratio  of  A.C.  to  B.C.  Voltage  in  a  Rotary  Converter. 

A  rotary  converter  resembles  in  its  general  construction  a  direct- 
current  machine,  except  that  the  armature  winding  is  connected 
not  only  to  the  commutator,  but  also  to  two  or  more  slip  rings.1 
When  such  a  machine  is  driven  mechanically  it  can  supply  a  direct 
current  through  its  commutator,  and  at  the  same  time  an  alter- 
nating current  through  its  slip  rings.  It  is  then  called  a  double- 
current  generator.  But  if  the  same  machine  is  connected  to  a 
source  of  alternating  voltage  and  brought  up  to  synchronous 
speed  it  runs  as  a  synchronous  motor  and  can  supply  direct  current 
through  its  commutator.  It  is  then  called  a  rotary  converter.  It 
is  also  sometimes  used  for  converting  direct  current  into  alter- 
nating current,  and  is  then  called  an  inverted  rotary. 

Both  the  direct  and  the  alternating  voltages  are  induced  in  the 
armature  of  a  rotary  converter  by  the  same  field,  and  our  problem 
is  to  find  the  ratio  between  the  two  voltages  for  a  given  arrange- 
ment of  the  slip  rings.  Consider  first  the  simplest  case  of  a  single- 
phase  converter  with  two  collector  rings  connected  to  the  arma- 
ture winding,  at  some  two  points  180  electrical  degrees'  apart.  If 
the  armature  has  a  multiple  winding  each  slip  ring  is  connected 
to  the  armature  in  as  many  places  as  there  are  pairs  of  poles.  In 
the  case  of  a  two-circuit  winding  each  collector  ring  is  connected 
to  the  armature  in  one  place  only. 

If  the  machine  has  p  poles  then  p  times  during  each  revolution 
the  direct-current  brushes  make  a  connection  with  the  same  arma- 
ture conductors  to  which  the  slip  rings  are  connected.  At  these 
moments  the  alternating  voltage  is  a  maximum,  because  the  direct- 
current  brushes  are  placed  in  the  position  where  the  induced  volt- 
age in  the  armature  is  a  maximum.  Thus,  with  two  slip  rings, 
connected  180  electrical  degrees  apart,  the  maximum  value  of  the 
alternating  voltage  is  equal  to  the  voltage  on  the  direct-current 
side.  If  the  pole  shoes  are  shaped  so  that  the  alternating  voltage 
is  approximately  sinusoidal,  the  effective  value  of  the  voltage 
between  the  slip  rings  is  I/ /v/2  =  70.7  per  cent  of  that  between  the 

1  See  the  author's  Experimental  Electrical  Engineering,  Vol.  2,  Chapter  28. 


78  THE  MAGNETIC  CIRCUIT  [ART.  32 

direct-current  brushes.  The  same  ratio  holds  true  for  a  two- 
phase  rota,ry,  for  the  voltages  induced  in  each  phase. 

Let  now  two  slip  rings  be  connected  at  two  points  of  the  arma- 
ture winding,  a  electrical  degrees  apart.     In  order  to  obtain  the 

value    of    the    alternating 
voltage  the  vectors  of  the 
voltages    induced    in    the 
individual    coils    must    be 
added  geometrically,  as  in 
Fig.  18.     With  a  large  num- 
ber of  coils  the  chords  can  be 
FiG.21.-Relationbetweenthealternating     ^placed  by  the  arc,  and  in 
voltages  in  a  rotary  converter.  tms  Wa7  FlS-  21  1S  obtained. 

The  diameter  MN  =  ei  of  the 

semicircle  represents  the  vector  of  the  alternating  voltage  when 
the  points  of  connection  to  the  slip  rings  are  displaced  by  180 
electrical  degrees,  while  the  chord  MP=ea  gives  the  voltage 
between  two  slip  rings  when  the  taps  are  distant  by  a  electrical 
degrees.  It  will  thus  be  seen  that 


(39) 


But  we  have  seen  before  that  ei  =  Q.7Q7E,  where  E  is  the  voltage 
on  the  direct-current  side  of  the  machine.  Hence,  for  sinusoidal 
voltages, 

ea=  0.707#  sin  Ja  .......     (40) 

The  following  table  has  been  calculated,  using  this  formula. 

Number  of  slip  rings  .......................       2          3        4       5          6 

Angle  between  the  adjacent  taps  in  electrical 

degrees  ................................   180       120      90     72        60 

Ratio  of  alternating  to  continuous  voltage,  in 

percent  ...............................  70.7     61.2     50     41.5     35.3 

The  foregoing  theory  shows  that  the  ratio  of  the  continuous  to 
the  alternating  voltage  is  fixed  in  a  given  converter,  and  in  order 
to  raise  the  value  of  the  direct  voltage  it  is  necessary  to  raise  the 
applied  alternating  voltage.  This  is  done  in  practice  either  by 
means  of  various  voltage  regulators  separate  from  the  converter, 
or  by  means  of  a  booster  built  as  a  part  of  the  converter.  Another 
method  of  varying  the  voltage  is  by  using  the  so-called  split-pole 
converter.  In  this  machine  the  distribution  of  the  flux  density  in 


CHAP.  IV]  INDUCED  E.M.F.  79 

the  air-gap  can  be  varied  within  wide  limits,  and  consequently 
that  component  of  the  field  which  is  sinusoidal  can  be  varied.  The 
result  is  that  the  ratio  of  the  direct  to  the  alternating  voltage  is 
also  variable.  Namely,  we  have  seen  before  that  the  value  of  the 
continuous  voltage  does  not  depend  upon  the  distribution  of  the 
flux,  but  only  upon  its  total  value,  while  the  effective  value  of  the 
alternating  voltage  depends  upon  the  sine  wave  or  fundamental 
component  of  the  flux  distribution.1 

Prob.  42.   Check  some  of  the  values  given  in  the  table  above. 

Prob.  43.  A  three-phase  rotary  converter  must  deliver  direct  current 
at  550  v.  What  is  the  voltage  on  the  alternating-current  side? 

Ans.     337  v. 

Prob.  44.  The  same  rotary  is  to  be  tapped  in  three  additional  places  so 
as  to  get  two-phase  current  also.  How  many  different  voltages  are  on 
the  alternating-current  side  and  what  are  they? 

Ans.    389,  337,  275, 195, 100. 

Prob.  45.  The  table  given  above  holds  true  only  when  the  flux  density 
is  distributed  approximately  according  to  the  sine  law.  Show  how  to 
determine  the  ratio  of  alternating  to  continuous  voltage  in  the  case  of  two 
collector  rings  connected  to  taps  180  electrical  degrees  apart,  when  the 
curve  of  field  distribution  is  given  graphically.  Solution :  Divide  the  pole 
pitch  into  a  sufficient  number  of  equal  parts  and  mark  them  on  a  strip  of 
paper.  Place  the  strip  along  the  axis  of  abscissae.  The  sum  of  the  ordi- 
nates  of  the  flux-density  curve,  corresponding  to  the  points  of  division,  at 
a  certain  position  of  the  strip,  gives  the  instantaneous  value  of  the  alter- 
nating voltage.  Having  performed  the  summation  for  a  sufficient  number 
of  positions  of  the  strip,  the  wave  of  the  induced  e.m.f.  is  plotted.  The 
scale  of  the  curve  is  determined  by  the  condition  that  the  maximum  ordi- 
nate  is  equal  to  the  value  of  the  continuous  voltage.  The  effective  value  is 
found  in  the  well-known  way,  either  in  rectangular  or  in  polar  coordinates 
(see  the  Electric  Circuit} . 

Prob.  46.  Apply  the  solution  of  the  preceding  problem  to  the  field  dis- 
tribution specified  in  Prob.  30,  Art.  30.  Ans.  81.5  per  cent. 

Prob.  47.  Extend  the  method  described  in  Prob.  45  to  the  case  when 
the  distance  between  the  taps  connected  to  the  slip  rings  is  less  than 
180  degrees.  Show  how  to  find  the  scale  of  voltage. 

Prob.  48.  How  does  a  fractional  pitch  affect  the  values  given  in  the 
table  above,  and  the  solution  outlined  in  Prob.  45? 

Prob.  49.  Show  how  to  solve  problems  45  to  48  when  the  field  curve 
is  given  analytically,  as  B  =  F(a),  for  instance  in  the  form  of  a  Fourier 
series.  Hint:  See  C.  A.  Adams,  "  Voltage  Ratio  in  Synchronous  Conver- 
ters with  Special  Reference  to  the  Split-pole  Converter,"  Trans.  Amer. 
Inst.  Elec.  Engs.,  Vol.  27  (1908),  part  II,  p.  959. 

1  See  C.  W.  Stone,  "  Some  Developments  in  Synchronous  Converters," 
Trans.  Amer.  Inst.  Elec.  Engs.,  Vol.  27  (1908),  p.  181. 


CHAPTER  V. 

THE   EXCITING   AMPERE-TURNS   IN    ELECTRICAL 
MACHINERY 

33.  The  Exciting  Current  in  a  Transformer.  The  magnetic 
flux  in  the  core  of  a  constant-potential  transformer  is  determined 
essentially  by  the  primary  applied  voltage,  and  is  practically 
independent  of  the  load  (see  Art.  25).  When  the  terminal 
voltage  is  given,  the  flux  becomes  definite  as  well.  The  ampere- 
turns  necessary  for  producing  the  flux  are  called  the  mag- 
netizing or  the  exciting  ampere-turns.  When  the  secondary  cir- 
cuit is  open  the  only  current  which  flows  through  the  primary 
winding  is  that  necessary  for  producing  the  flux.  This  current  is 
called  the  no-load,  exciting,  or  magnetizing  current  of  the  trans- 
former. When  the  transformer  is  loaded,  the  vector  difference 
between  the  primary  and  the  secondary  ampere-turns  is  practi- 
cally equal  to  the  exciting  ampere-turns  at  no-load. 

The  exciting  current  is  partly  reactive,  being  due  to  the 
periodic  transfer  of  energy  between  the  el'ectric  and  the  magnetic 
circuits  (see  Art.  16  above),  partly  it  represents  a  loss  of  energy 
due  to  hysteresis  and  eddy  currents  in  the  core.  Some  writers  call 
the  reactive  component  of  the  no-load  current  the  magnetizing 
current,  and  the  total  no-load  current  the  exciting  current. 
Generally,  however,  the  words  magnetizing  and  exciting  are  used 
interchangeably  to  denote  the  total  no-load  current.  The 
components  of  the  current  in  phase  and  in  quadrature  with  the 
induced  voltage  are  called  the  energy  and  the  reactive  components 
respectively. 

The  no-load  or  exciting  current  in  a  transformer  must  usually 
not  exceed  a  specified  percentage  of  the  rated  full-load  current ; 
it  is  therefore  of  importance  to  know  how  to  calculate  the 
exciting  current  from  the  given  dimensions  of  a  transformer. 
Knowing  the  applied  voltage  and  the  number  of  turns,  the  maxi- 
mum value  of  the  flux  is  calculated  from  eq.  (28).  We  shall 

80 


CHAP.  V]  EXCITING  AMPERE-TURNS  81 

assume  first  that  the  maximum  flux  density  in  the  core  is  so 
low,  that  it  lies  practically  on  the  straight  part  of  the  mag- 
netization curve  of  the  material  (Fig.  3).  The  case  of  high  flux 
densities  is  considered  in  the  next  article. 

Since  by  assumption  the  instantaneous  magnetomotive  forces 
are  proportional  to  the  corresponding  flux  densities,  the  magnetiz- 
ing current  must  vary  according  to  the  sine  law.  It  is  sufficient, 
therefore,  to  calculate  the  maximum  value  of  the  magnetomotive 
force,  corresponding  to  the  maximum  flux.  Knowing  the  ampli- 
tude ®m  of  the  flux  and  the  net  cross-section  of  the  core,  A,  the 
flux  density  Bm  becomes  known  ;  from  the  magnetization  curve  of 
the  material  (Fig.  3)  the  corresponding  value  of  Hm,  or  the  ampere- 
turns  per  unit  length  of  path,  is  found.  The  mean  length  I  of  the 
lines  of  force  is  determined  from  the  drawing  of  the  core,  so  that 
the  total  magnetizing  ampere-turns  Mm  =  Hml  can  be  calculated. 
The  mean  magnetic  path  around  the  corners  is  somewhat  shorter 
than  the  mean  geometric  path. 

Let  HI  be  the  number  of  turns  in  the  primary  winding,  and  i0 
the  effective  value  of  the  reactive  component  of  the  exciting  cur- 
rent. We  have  then 

2  =  Mm  ........     (41) 


From  this  equation  the  quantity  which  is  unknown  can  be  calcu- 
lated. 

It  is  presupposed  in  the  above  deduction  that  the  joints 
between  the  laminations  offer  no  reluctance.  In  reality, 
the  contact  reluctance  is  appreciable;  its  value  depends  upon 
the  character  of  the  joints,  and  the  care  exercised  in  the 
assembling  of  the  core.  This  reluctance  of  the  joints  can  be 
expressed  by  the  length  of  an  equivalent  air-gap  having  the  same 
reluctance.  Thus,  experiments  show  that  each  overlapping  joint 
is  equivalent  to  an  air-gap  0.04  mm.  long.  A  butt  joint,  with 
very  careful  workmanship,  is  equivalent  to  an  air-gap  of  about 
0.05  mm.;  in  practice,  a  butt  joint  may  offer  a  reluctance  of  from 
50  to  100  per  cent  higher  than  the  foregoing  value.1  Knowing  the 

1  H.  Bohle,  "Magnetic  Reluctance  of  Joints  in  Transforming  Iron,"  Journal 
(British)  Inst.  Electr.  Engs,,  Vol.  41,  1908,  p.  527.  It  is  convenient  to 
estimate  the  influence  of  the  joints  in  ampere-turns  at  a  standard  flux 
density.  For  each  lap  joint  32  ampere-turns  must  be  added  at  a  density 
of  10  kilolines  per  square  centimeter,  while  a  butt  joint  requires  at  the  same 


82  THE  MAGNETIC  CIRCUIT  [ART.  33 

length  a  of  the  equivalent  air-gap,  the  number  of  additional  ampere- 
turns  is  calculated  according  to  the  formula  aBm/  /*,  and  then  is 
multiplied  by  the  number  of  joints  in  series  (usually  four).  This 
number  of  ampere-turns  must  be  added  to  Mm  calculated  above. 
The  energy  component  i\  of  the  exciting  current  is  determined 
from  the  power  lost  in  hysteresis  and  eddy  currents  in  the  core. 
Having  calculated  this  power  P  as  is  explained  in  Article  19,  we 
find  ii  =  P/Ei,  where  E\  is  the  primary  applied  voltage.  Knowing 
i0  and  ii,  the  total  no-load  current  is  found  as  their  geometric  sum, 


The  watts  expended  in  core  loss  depend  only  upon  the  volume 
of  the  iron,  the  frequency,  and  the  flux  density  used.  It  can  be 
also  shown  that  the  reactive  volt-amperes  required  for  the  excita- 
tion of  the  magnetic  circuit  of  a  transformer  depend  only  upon  the 
volume  of  the  iron,  the  frequency,  and  the  flux  density.  Namely, 
neglecting  the  influence  of  the  joints,  eq.  (41)  can  be  written  in  the 
form 


Eq.  (28)  in  Art.  25  can  be  written  as 


where  A  is  the  cross-section  of  the  iron,  and  Bm  is  the  maximum 
flux  density,  in  kilolines  per  square  centimeter.  Multiplying  these 
two  equations  together,  term  by  term,  and  cancelling  HI  we  get, 
after  reduction, 

Eiio/V-nfBnHnXlQ'*,        ....   .      (42) 

where  V=Al  is  the  volume  of  the  iron,  in  cubic  centimeters. 
The  left-hand  side  of  eq.  (42)  represents  the  reactive  magnetizing 
volt-amperes  per  unit  volume  of  iron;  the  right-hand  side  is  a 
function  of  /  and  Bm  only,  because  Hm  can  be  expressed  through 
Bm  from  the  magnetization  curve  of  the  material. 

Formula  (42)  can  be  plotted  as  a  set  of  curves,  one  for  each 
commercial  frequency.  These  curves  are  quite  convenient  in  the 
design  of  transformers,  because  they  enable  one  to  estimate 
directly  either  the  permissible  volume  of  iron,  or  the  permissible 
flux  density,  when  the  reactive  component  of  the  exciting  cur- 

density  from  60  to  80  ampere-turns.  At  other  flux  densities  the  increase 
is  proportional. 


CHAP.  V]  EXCITING  AMPERE-TURNS  83 

rent  is  limited  to  a  certain  percentage  of  the  full -load  current.  In 
practice,  such  curves  are  sometimes  plotted  directly  from  the 
results  of  tests  on  previously  built  transformers.  These  experi- 
mental curves  are  the  most  secure  guide  for  predicting  the  exciting 
current  in  transformers;  formula  (42)  shows  their  rational  basis. 

Prob.  1.  Prove  that  if  there  were  no  core  loss  the  exciting  current 
would  be  purely  reactive,  that  is  to  say,  in  a  lagging  phase  quadrature 
with  the  induced  voltage. 

Prob.  2.  The  core  of  a  22-kv.  25-cycle  transformer,  like  the  one 
shown  in  Fig.  12,  has  a  gross  cross-section  of  4500  sq.cm.;  the  mean 
path  of  the  lines  of  force  is  420  cm.;  the  material  is  silicon  steel;  the 
maximum  flux  is  36  megalines.  The  expected  reluctance  of  each  of  the 
four  butt  joints  is  estimated  to  be  equivalent  to  an  0.08  mm.  air-gap. 
What  are  the  two  components  of  the  exciting  current,  and  what  is  the 
total  no-load  current?  Ans.  1.8;  0.4;  1.85. 

Prob.  3.  In  what  respects  does  the  calculation  of  the  magnetizing 
current  in  a  shell-type  or  cruciform-type  transformer  differ  from  that 
in  a  core-type  transformer? 

Prob.  4.  Show  that  for  flux  densities  up  to  10  kl. /sq.cm.  the  mag- 
netizing volt-amperes  per  kilogram  of  carbon  steel  at  60  cycles  are 
approximately  equal  to  7.3  (5m/ 10) 2. 

Prob.  5.  Show  that  the  influence  of  the  joints  can  be  taken  into 
account  in  formula  (42)  by  adding  to  the  actual  volume  of  the  iron 
the  volume  of  the  air-gaps  multiplied  by  the  relative  permeability  of 
the  iron. 

Prob.  6.  A  shell-type  1000-kva.,  60-cycle  transformer  is  to  have  a 
core  made  of  silicon-steel  punchings  of  a  width  w  =  l7  cm.  (Fig.  13); 
the  average  length  of  the  magnetic  path  in  iron  is  180  cm. ;  the  reactive 
component  of  the  no-load  current  must  not  exceed  2  per  cent  of  the 
full-load  current.  Draw  curves  of  the  required  height  of  the  core  per 
link,  and  of  the  total  core  loss  in  per  cent  of  the  rated  kva.,  for  flux 
densities  up  to  10  kl. /sq.cm. 

Ans.  £2/i=5200;  at  5=9,  P=0.51  per  cent. 

34.  The  Exciting  Current  in  a  Transformer  with  a  Saturated 
Core.  In  the  preceding  article  the  flux  density  in  the  core  is 
supposed  to  be  within  the  range  of  the  straight  part  of  the  satura- 
tion curves  (Fig.  3),  so  that,  when  the  flux  varies  according- to  the 
sine  law,  the  magnetizing  current  also  follows,  a  sine  wave.  We 
shall  now  Consider  the  case  when  the  flux  density  rises  to  a  value 
on  or  beyond  the  knee  of  the  magnetization  curve.  Such  high  flux 
densities  are  used  with  silicon  steel  cores,  especially  at  low  frequen- 
cies. In  this  case  the  magnetizing  current  does  not  vary  according 
to  the  sine  law,  but  is  a  peaked  wave,  because  at  the  moments  when 


84 


THE   MAGNETIC  CIRCUIT 


[ART.  34 


the  flux  is  approaching  its  maximum,  the  current  is  increasing 
faster  than  the  flux,  on  account  of  saturation.  The  amplitude 
factor  of  the  current  wave,  or  the  ratio  of  the  amplitude  to  the 
effective  value  is  no  more  equal  to  \/2,  but  is  larger.  Let  this 
ratio  be  denotedby  /a.  Then  eq.  (41)  becomes 


(43) 


where  to  is  as  before  the  effective  value  of  the  reactive  component 
of  the  exciting  current.     The  value  of  %a  is  obtained  by  actually 


3.0 


2.5 


.2.0 


il.5 


1.0 


0.5 


0.0 


10 


15 
Flux  density  in  Kl.  per  Sq.  Cm. 


FIG.  22. — Ratio  of  the  amplitude  to  the  effective  value  of  the  magnetizing 

current. 

plotting  the  curve  of  the  magnetizing  current  from  point  to  point 
and  calculating  its  effective  value.  Since  the  procedure  is  rather 
long,  it  is  convenient  to  calculate  the  values  of  #a  once  for  all  for 
the  working  range  of  values  Bm.  This  has  been  done  for  the  mate- 
rials represented  in  Fig.  3,  and  the  results  are  plotted  in  Fig.  22. 

Strictly  speaking,  the  exciting  current  is  unsymmetrical,  due 
to  the  effect  of  hysteresis,  and  the  values  of  ya  ought  to  be  cal- 
culated, using  the  hysteresis  loops  of  the  steel.  However,  it  is  very 
nearly  correct  to  calculate  %a  from  the  magnetization  curve,  and  to 


CHAP.  V]  EXCITING  AMPERE-TURNS  85 

calculate  the  energy  component  of  the  exciting  current  separately, 
from  the  core  loss  curves  (Fig.  10).  The  magnetizing  current 
required  for  the  joints  is  calculated  separately,  using  ya=  \/2, 
according  to  eq.  (41).  The  total  effective  magnetizing  current  is 
found  by  adding  together  the  values  of  IQ  for  the  iron  and  for  the 
joints.  The  loss  component,  i\,  is  added  to  this  value  in  quadra- 
ture, to  get  the  total  no-load  current.  As  is  mentioned  above,  it 
is  preferred  in  practice  to  estimate  the  total  exciting  current  of  new 
transformers  from  the  curves  of  no-load  volt-amperes  per  kilogram 
of  iron,  the  values  being  obtained  from  tests  on  similar  trans- 
formers. 

Prob.  7.  The  core  of  a  25-cycle  cruciform  type  transformer  (Fig.  14) 
weighs  265  kg.;  the  mean  length  of  the  magnetic  path  is  170  cm.;  the 
material  is  silicon  steel.  The  4400-v.  winding  of  the  transformer  has 
1100  turns  in  series.  What  is  the  reactive  component  of  the  no-load 
current?  Ans.  8.4  amperes. 

Prob.  8.  Check  a  few  points  on  the  curves  in  Fig.  22. 

Prob.  9.  Show  that  in  formula  (42)  the  coefficient  n  is  a  special 
case  of  the  more  general  factor  4.44/£a,  when  the  magnetizing  current 
does  not  follow  the  sine  wave. 

Prob.  10.  What  are  the  reactive  volt-amperes  per  kilogram  of  carbon 
steel  at  40  cycles  and  at  a  flux  density  of  16  kl./sq.cm.?  Ans.  56.4. 

Prob.  11.  Show  how  to  calculate  the  exciting  ampere-turns  required 
for  a  given  flux  in  a  thick  and  short  core  in  which  the  flux  density  is 
different  along  different  paths. 

35.  The  Types  of  Magnetic  Circuit  Occurring  in  Revolving 
Machinery.  The  remainder  of  this  chapter  and  the  next  chapter 
have  for  their  object  the  calculation  of  the  exciting  ampere-turns 
necessary  for  producing  a  certain  useful  flux  in  the  principal  types 
of  electric  generators  and  motors.  In  direct-current  machines, 
in  alternators,  and  in  rotary  converters  it  is  necessary  to  know  the 
exciting  or  field  ampere-turns  in  order  to  plot  the  no-load  satura- 
tion curve,  to  predict  the  performance  of  the  machine  under  vari- 
ous loads,  and  to  design  the  field  coils.  In  an  induction  motor  one 
wants  to  know  the  required  excitation  in  order  to  determine  the 
no-load  current,  or  to  calculate  the  number  of  turns  in  the  stator 
winding,  when  the  limiting  value  of  the  no-load  current  is  pre- 
scribed. The  general  procedure  in  determining  the  required 
number  of  ampere-turns  for  a  given  flux  is  in  many  respects  the 
same  in  all  the  types  of  electrical  machinery,  so  that  it  is  possible 
to  outline  the  general  method  before  going  into  details. 


86 


THE  MAGNETIC  CIRCUIT 


.[ART.  35 


In  direct-current  machines  and  in  synchronous  generators, 
motors,  and  rotary  converters,  the  magnetic  flux  (Figs.  15  and  20) 
from  a  field  pole  passes  into  the  air-gap  and  the  armature  teeth. 
In  the  armature  core  the  flux  is  divided  into  two  halves,  each  half 
going  to  one  of  the  adjacent  poles.  The  magnetic  paths  are  com- 
pleted through  the  field  frame.  Part  of  the  flux  passes  directly 
from  one  pole  to  the  two  adjacent  poles  through  the  air,  without 
going  through  the  armature.  This  part  of  the  flux  is  known  as 
the  leakage  flux.  The  closed  magnetic  paths  and  the  field  coils  of  a 
machine  may  be  thought  of  as  the  consecutive  links  of  a  closed 
chain.  While  in  a  transformer  the  chain  is  open,  in  generators 


FIG.  23. — The  paths  of  the   main  flux  and  of   the   leakage  fluxes  in  an 
induction  motor  (or  generator). 

and  motors  the  chain  must  be  closed  on  account  of  the  continuous 
rotation. 

In  induction  machines,  both  generators  and  motors  (Fig.  23), 
the  flux  at  no  load  is  produced  by  the  currents  in  the  stator  wind- 
ings only.  When  the  machine  is  loaded,  the  flux  is  produced  by 
the  combined  action  of  the  stator  and  rotor  currents,  the  rotor  cur- 
rents opposing  those  in  the  stator,  the  same  as  in  a  transformer. 
Therefore,  the  flux  in  the  loaded  machine  may  be  regarded  as  the 
resultant  of  the  following  three  component  fluxes:  The  main  or 
useful  flux,  $,  which  links  with  both  the  primary  and  the  secondary 
windings;  the  primary  leakage  flux,  $1;  which  links  with  the  stator 
winding  only ;  and  the  secondary  leakage  flux,  $2  which  is  linked 
with  the  rotor  winding  alone.  The  leakage  fluxes  not  only  do  not 


CHAP.  V]  EXCITING  AMPERE-TURNS  87 

contribute  to  the  useful  torque  of  the  machine,  but  actually  reduce 
it.  In  reality,  there  is  of  course  but  one  flux,  the  resultant  of  the 
three,  but  for  the  purposes  of  theory  and  computations  the  three 
component  fluxes  can  be  considered  as  if  they  had  a  real  separate 
existence.  In  this  and  in  the  following  chapter  the  main  flux 
only  will  be  discussed  for  this  type  of  machinery.  Considera- 
tion of  the  leakage  flux  will  be  reserved  to  Art.  66. 

The  total  magnetomotive  force  per  magnetic  circuit  is  equal 
to  the  sum  of  the  m.m.fs.  necessary  for  establishing  the  required 
flux  in  the  separate  parts  of  the  circuit  which  are  in  series,  viz.,  the 
pole-pieces,  the  air-gap,  the  teeth,  and  the  armature  core.  All  the 
necessary  elements  for  the  solution  of  this  problem  have  been  dis- 
cussed in  the  first  two  chapters.  It  remains  here  to  establish  some 
semi-empirical  "  short-cut  "  rules  and  formulae  for  the  irregular 
parts  of  the  circuit,  for  which,  although  close  approximations  can 
be  made,  the  exact  solution  is  either  impossible  or  too  complicated 
for  the  purposes  of  this  text.  The  following  topics  are  considered 
more  in  detail  in  the  subsequent  articles  of  this  and  of  the  follow- 
ing chapter. 

(a)  The  ampere-turns  necessary  for  the  air-gap  when  it  is 
limited  on  one  side  or  on  both  sides  by  teeth,  so  that  the  flux  den- 
sity in  the  air-gap  is  not  uniform. 

(6)  The  ampere-turns  necessary  for  the  armature  teeth  when 
they  are  so  highly  saturated  that  an  appreciable  part  of  the  flux 
passes  through  the  slots  between  the  teeth. 

(c)  The  ampere-turns  necessary  for  the  highly  saturated  cores 
in  which  the  lengths  of  the  individual  paths  differ  considerably 
from  one  another,  with  a  consequent  lack  of  uniformity  in  the  flux 
density. 

(d)  The  leakage  coefficient  and  the  value  of  the  leakage  flux 
which   passes   directly   from   pole   to   pole.     This   leakage    flux 
increases  the  flux  density  in  the  poles  and  in  the  field  frame  of  the 
machine,    and   consequently   increases   the   required   number   of 
ampere-turns. 

All  of  the  m.m.f.  calculations  that  follow  are  per  pole  of  the 
machine,  or  what  is  the  same,  for  one-half  of  a  complete  magnetic 
circuit  (cdfg  in  Figs.  15,  20,  and  23),  the  two  halves  being  identical. 
This  fact  must  be  borne  in  mind  when  comparing  the  formulae  with 
those  given  in  other  books,  in  which  the  required  ampere-turns  are 
sometimes  calculated  for  a  complete  magnetic  circuit. 


88 


THE  MAGNETIC  CIRCUIT 


[ART.  36 


Prob.  12.  Inspect  working  drawings  of  electrical  machines  found 
in  various  books  and  magazine  articles;  indicate  the  paths  of  the  main 
and  of  the  leakage  fluxes;  and  make  clear  to  yourself  the  reasons  for  the 
use  of  different  kinds  of  steel  and  iron  in  the  frame,  the  core,  the  pole- 
pieces,  and  the  pole  shoes. 

Prob.  13.  Make  sketches  of  the  magnetic  circuit  of  a  turbo-alternator 
with  a  distributed  field  winding,  of  a  homopolar  machine,  of  an  inductor- 
type  alternator,  and  of  a  single-phase  commutator  motor.  Indicate  the 
paths  of  the  useful  and  of  the  leakage  fluxes. 

36.  The  Air-gap  Ampere  Turns.  The  general  character  of  the 
distribution  of  the  magnetic  flux  in  the  air-gap  of  a  synchronous 
and  of  a  direct-current  machine  is  shown  in  Fig.  24,  the  curvature 
of  the  armature  being  disregarded.  The  principal  features  of  this 
flux  distribution  are  as  follows : 


Armature 

4-X-4 


^Air-duct 


mature 


FIG.  24. — The  cross-section   of   a   direct-current   or  synchronous  machine, 
showing  the  flux  in  the  air-gap. 

(a)  The  flux  per  tooth  pitch  A  is  practically  the  same  under  all 
the  teeth  in  the  middle  part  of  the  pole,  where  the  air-gap  has  a 
constant  length,  and  is  smaller  for  the  teeth  near  the  pole-tips 
where  the  air-gap  is  larger. 

(b)  On  the  armature  surface  the  flux  is  concentrated  mainly 
at  the  tooth-tips;    very  few  lines  of  force  enter  the  armature 
through  the  sides  and  the  bottom  of  the  slots. 

(c)  There  is  a  considerable  spreading,  or  fringing,  of  the  lines 
of  force  at  the  pole-tips. 

(d)  In  the  planes  passing  through  the  axis  of  the  shaft  of  the 
machine  there  is  also  some  spreading  or  fringing  of  the  lines  of 
force  at  the  flank  surfaces  of  the  armature  and  the  pole,  and  in 
the  ventilating  ducts. 


CHAP.  V]  EXCITING  AMPERE-TURNS  89 

This  picture  of  the  flux  distribution  follows  directly  from  the 
fundamental  law  of  the  magnetic  circuit,  the  flux  density  being 
higher  at  the  places  where  the  permeance  of  the  path  is  higher. 
The  actual  flux  distribution  is  such  that  the  total  permeance  of  all 
the  paths  is  a  maximum,  as  compared  to  any  other  possible  distri- 
bution. In  other  other  words,  the  flux  distributes  itself  in  such  a 
way,  that  with  a  given  m.m.f  .  the  total  flux  is  a  maximum,  or  with 
a  given  flux  the  required  m.m.f.  is  a  minimum.  This  is  confirmed 
by  the  beautiful  experiments  of  Professor  Hele-Shaw  and  his  col- 
laborators,1 who  have  obtained  photographs  of  the  stream  lines  of  a 
fluid  flowing  through  an  arrangement  which  imitated  the  shape 
and  the  relative  permeances  of  the  air-gap  and  of  the  teeth  in  an 
electric  machine. 

Let  (Pa  be  the  total  permeance  in  perms  of  the  air-gap  between 
the  surface  of  the  pole  shoe  and  the  teeth,  and  let  0  be  the  useful 
flux  per  pole,  in  maxwells,  which  is  supposed  to  be  given.  Then, 
according  to  eq.  (2),  Art.  5,  the  number  of  ampere-turns  required 
for  the  air-gap  is 


(44) 


The  problem  is  to  calculate  the  permeance  of  the  gap  from  the 
drawing  of  the  machine. 

One  of  the  usual  practical  methods  is  to  calculate  (P  under  cer- 
tain simplifying  assumptions  and  then  multiply  the  result  by  an 
empirical  coefficient  determined  from  tests  on  similar  machines. 
The  simplest  assumptions  are  (Fig.  25)  :  (a)  that  the  armature  has 
a  smooth  surface,  the  slots  being  filled  with  iron  of  the  same  per- 
meability as  that  of  the  teeth;  (6)  that  the  external  surface  of  the 
pole  shoes  is  concentric  with  that  of  the  armature;  (c)  that  the 
equivalent  air-gap  aeq  is  equal  to  two-thirds  of  the  minimum  air- 
gap  plus  one-third  of  the  maximum  air-gap  of  the  actual  machine  ; 
(d)  that  the  ventilating  ducts  are  filled  with  iron  ;  (e)  that  the  paths 
of  the  fringing  flux  at  the  edges  of  the  pole  shoe  are  straight  lines, 
and  extend  longitudinally  to  the  edge  of  the  armature  surface  and 
laterally  for  a  distance  equal  to  the  equivalent  air-gap  on  each 
side. 

1  For  a  detailed  account  of  the  experimental  and  theoretical  investigations 
on  this  subject,  with  numerous  references,  see  Hawkins  and  Wallis,  The 
Dynamo  (1909),  Vol.  1,  Chapter  XV. 


90 


THE  MAGNETIC  CIRCUIT 


[ART.  36 


With  these  assumptions,  the  permeance  of  the  "  simplified  " 
air-gap  is 

......     (45) 


where  w8  is  the  average  width  of  the  flux,   and  ls  is  its  average 
axial  length.     Or 

ws  =  i  (wa  +  Wp)  =  wp  +  a'  ; 


The  lateral  spread  a!  of  the  lines  of  force  at  each  pole-tip  is 
taken  to  be  approximately  equal  to  a^. 


f 

1/t 

«> 

1 

i  1 

I 

II 

I 

! 

Mimm 

j/     a*<l 

t 

IMMMMMIM 

— lp ! 


FIG.  25.  —  Magnetic  flux  in  the  simplified  air-gap. 

The  permeance  of  the  actual  air-gap  is  smaller  than  that  of  the 
simplified  gap,  so  that  we  have 


.......      (46) 

where  ka  is  a  coefficient  larger  than  unity,  called  the  air-gap  factor. 
Substituting  the  value  of  (P  from  eq.  (46)  into  (44)  gives 


(47) 


so  that  ka  is  the  factor  by  which  the  ampere-turns  for  the  simplified 
air-gap  must  be  multiplied  in  order  to  obtain  the  ampere-turns 
required  for  the  actual  air-gap.  The  value  of  ka  usually  varies 
between  1.1  and  1.3,  depending  on  the  relative  proportions  of  the 
teeth,  the  slots,  and  the  air-gap,  and  on  the  shape  of  the  poles. 

The  numerical  values  of  fcaare  calculated  from  the  results  of  tests 
on  machines  of  proportions  similar  to  that  being  computed.  Let 
the  no-load  saturation  curve  of  a  machine  be  available  from  test ; 
this  is  a  curve  which  gives  the  relation  between  the  induced  voltage 


CHAP.  V]  EXCITING  AMPERE-TURNS  91 

and  the  field  current  of  the  machine.  From  the  known  specifica- 
tions of  the  machine  this  curve  can  be  easily  converted  into  one 
which  gives  the  useful  flux  per  pole  against  the  ampere-turns  per 
pole  as  abscissae.  The  lower  part  of  such  a  curve  is  always  a 
straight  line,  there  being  then  practically  no  saturation  in  the  iron. 
On  this  part  of  the  curve,  practically  the  whole  m.m.f .  is  consumed 
in  the  air-gap,  so  that  the  actual  permeance  of  the  air-gap  is  found 
by  dividing  one  of  the  ordinates  by  the  corresponding  abscissa. 
The  permeance  of  the  simplified  air-gap  is  calculated  from  eq.  (45), 
and  the  ratio  of  the  two  gives  the  value  of  the  coefficient  ka.  This 
value  is  then  used  in  the  design  and  calculation  of  the  performance 
of  new  machines  with  similar  proportions.  Engineering  judg- 
ment and  practical  experience  are  factors  of  considerable  impor- 
tance in  estimating  the  values  of  ka  for  new  machines. 

The  same  method  of  calculating  the  air-gap  ampere-turns  is 
applicable  to  induction  machines  (Fig.  23).  The  ampere-turns 
are  computed,  assuming  both  the  rotor  and  the  stator  to  have 
smooth  iron  surfaces,  without  slots;  the  result  is  then  multiplied 
by  a  factor  ka  larger  than  unity,  determined  from  tests  upon 
machines  of  similar  proportions. 

A  more  accurate,  though  more  elaborate,  method  for  calcula- 
ting the  air-gap  ampere-turns  is  explained  in  the  next  article. 

Prob.  14.  Calculate  the  air-gap  ampere-turns  per  pole  for  a  6600  v., 
25-cycle,  375-r.p.m.  alternator  to  be  built  according  to  the  following 
specifications:  The  bore  2.4  m.;  the  gross  axial  length  of  the  armature 
core  55  cm.;  seven  air  ducts  9  mm.  each;  the  minimum  air-gap  is  15 
mm.;  the  maximum  air-gap  is  30  mm.  The  poles  cover  66  per  cent  of 
the  periphery;  the  axial  length  of  the  pole  shoes  is  53  cm.  The  useful 
flux  per  pole  at  no-load  and  at  the  rated  voltage  is  19.1  megalines.  The 
air-gap  factor  is  estimated  to  be  about  1.15.  Ans.  10,600. 

Prob.  15.  The  no-load  characteristic  obtained  from  the  test  upon 
the  machine  specified  in  the  preceding  problem  has  a  straight  part  such 
that  at  a  field  current  of  52  amp.  the  line  voltage  is  4000  v.  Each  field 
coil  has  120  turns.  What  is  the  true  value  of  the  air-gap  factor? 

Ans.     1.12. 

Prob.  16.  A  pole  shoe  is  so  shaped  that  the  minimum  air-gap  is  a0 
and  the  maximum  air-gap  is  a1=a0+Ja,  the  increase  in  the  length  being 
proportional  to  the  square  of  the  distance  from  the  center  of  the  pole. 
What  is  the  length  aeq  of  the  equivalent  uniform. air-gap  such  that  its 
total  permeance  is  the  same  as  that  of  the  given  air-gap?  Assume  a 
smooth-body  armature,  and  neglect  the  fringing  at  the  pole-tips.  Solu- 
tion: Let  the  peripheral  width  of  the  pole  be  2w;  then  the  length  of 


92  THE   MAGNETIC  CIRCUIT  [ART.  37 

the  air-gap  at  a  distance  x  from  the  center  is  a  =  a0  +da(x/w}2.  The  per- 
meance of  an  infinitesimal  path  of  the  width  dx  is  proportional  to  dx/ax. 
Hence  we  have  the  relation 


x.w 

I     dx/[a0  +Aa(x/w)  2]=  w/aeq, 
JQ 


from  which 

aeq  =  Va^a/tan"1  (  \/Ja/a0)  . 

Prob.  17.  What  is  the  length  of  the  equivalent  air-gap  in  the  preced- 
ing problem  if  the  clearance  at  the  pole-tips  is  twice  the  clearance  at 
the  center  of  the  pole?  Ans.  1.273a0. 

Prob.  18.  Show  that,  when  the  air-gap  is  non-uniform,  the  length 
of  the  equivalent  uniform  gap  can  be  determined  approximately,  accord- 
ing to  Simpson's  Rule,  from  the  equation 


where  a0,  at,  and  am  are  the  lengths  of  the  gap  at  the  center,  at  the  tip  of 
the  pole,  and  midway  respectively.  If  the  air-gap  is  uniform  under  the 
major  portion  of  the  pole,  but  the  pole  shoe  is  chamfered,  more  terms 
must  be  taken  in  Simpson's  formula  in  order  to  obtain  aeq  with  a 
sufficient  accuracy. 

Prob.  19.  What  is  the  length  of  the  air-gap  required  in  problems 
16  and  17,  according  to  the  formula  given  in  problem  18? 

Ans.     1.276a0. 

37.  The  Method  of  Equivalent  Permeances  for  the  Calculation 
of  Air-gap  Ampere-turns.  An  inspection  of  Fig.  24  will  show  that 
the  total  permeance  of  the  air-gap  is  made  up  of  a  number  of  per- 
meances in  parallel.  It  is  equal  therefore  to  the  sum  of  these 
permeances.  For  the  purpose  of  calculation  two  kinds  of  per- 
meances are  considered  separately:  those  from  the  teeth  to 
the  pole  surface  proper,  and  those  from  the  teeth  to  the  pole-tips. 
The  former  can  be  calculated  quite  accurately,  the  latter  are  to 
some  extent  estimated. 

The  permeance  per  tooth  pitch  in  the  part  of  the  air-gap  near 
the  center  of  the  pole  can  be  divided  into  two  parts,  that  under  the 
tooth-tip,  and  the  fringe  from  the  sides  of  the  slots  and  in  the  ven- 
tilating ducts.  The  permeance  of  the  paths  which  proceed  from 
the  tooth-tip  constitutes  the  larger  portion  and  is  made  up  of 
nearly  parallel  lines;  this  permeance  is  therefore  easily  computed. 
The  values  of  the  permeance  of  the  fringe  from  the  flank  of  the 
tooth  to  the  perpendicular  surface  of  the  pole  have  been  deter- 


CHAP.  V]  EXCITING  AMPERE-TURNS  93 

mined  theoretically  by  Mr.  F.  W.  Garter.1  Only  the  numerical 
results  are  given  here,  in  a  somewhat  simplified  practical  form; 
the  solution  itself  presupposing  a  knowledge  of  the  properties  of 
conjugate  functions.2 

Consider  the  permeance  of  two  tooth  fringes,  such  as  opqr  and 
o'p'q'r'  (Fig.  24),  perpendicular  to  the  plane  of  the  paper.  This 
permeance  depends  only  upon  the  ratio  of  the  slot  width  s  to  the 
length  a  of  the  air-gap,  for  let  both  s  and  a  be  increased  say  twice : 
The  length  and  the  cross-section  of  each  elementary  tube  of  force 
is  also  increased  twice,  hence  its  permeance  remains  the  same. 

The  permeance  of  each  fringe  can  be  replaced  by  the  permeance 
of  an  equivalent  rectangular  path  of  the  length  a  and  of  a  width 
%At  (Fig.  26).  This  is  the  same  as  increasing  the  width  of  the 
tooth  by  the  amount  At  and  assuming  all  the  lines  of  force  to  be 
parallel  to  each  other  in  the  air-gap.  The  permeance  of  the  path 
which  replaces  the  two  fringes  is  equal  to  pAt/a.  From  what  has 
been  said  above  follows  that  the  ratio  At /a  depends  only  upon  the 
ratio  of  s/a;  the  relationship  between  the  two  ratios  is  plotted  in 
Fig.  26,  from  Carter's  calculations.  For  the  sake  of  convenience 
and  accuracy,  the  curve  is  drawn  to  two  different  scales,  one  for 
large  the  other  for  small  values  of  s/a. 

The  curve  in  Fig.  26  may  be  interpreted  in  two  ways:  It  may 
be  said  to  represent  the  "geometric  permeance''  of  the  fringe  (for 
/*=!);  or  else  it  may  be  said  to  give  the  correspondings  sets  of 
values  of  s  and  At,  measured  in  the  lengths  of  the  air-gap  as  the 
unit.  With  a  given  a,  At  increases  with  s,  because  the  maximum 
width  of  the  actual  fringe  is  Js.  With  a  given  s  the  width  At 
increases  toward  the  pole-tip  (if  the  air-gap  is  variable),  because 
with  a  longer  air-gap  the  fringing  lines  of  flux  fill  a  larger  part  of 
the  air-gap  under  the  slot. 

The  corrected  width  of  the  tooth  is  t'  =  t+ At  and  the  permeance 
of  the  air-gap,  in  perms  per  tooth  pitch,  is 

<Pat=135(M/ax+t/ax)lrt,       ....     (48) 

^ote  on  Air-gap  Induction,  Journ.  Inst.  Electr.  Eng.  (British),  Vol.  29, 
(1899-1900),  p.  929;  Air-gap  Induction,  Electrical  World,  Vol.  38,  (1901) 
p.  884;  See  also  Hawkins  and  Wallis,  The  Dynamo  (1909),  Vol.  1,  p.  446; 
E.  Arnold,  Die  Gleichstrommaschine  (1906),  Vol.  1,  p.^266. 

2  J.  C.  Maxwell,  Electricity  and  Magnetism,  Vol.  1,  p.  284;  J.  J.  Thomson, 
Recent  Researches  in  Electricity  and  Magnetism,  Chapter  III;  Horace  Lamb, 
Hydrodynamics  (1895),  Chapter  IV. 


94  THE  MAGNETIC  CIRCUIT  [ART.  37 

where  ax  is  the  length  of  the  air-gap  at  the  center  of  the  tooth,  and 
leff  is  the  effective  axial  length  of  the  machine  (see  below) .  The 
value  of  At/ax  must  be  taken  from  Fig.  26  for  the  corresponding 
ratio  s/ax. 

The  permeance  of  the  pole  fringe,  hmn  in  Fig.  24,  cannot  be  cal- 
culated by  the  foregoing  method,  because  this  permeance  depends 
upon  the  irregular  shape  of  the  pole-tip.  The  pole-fringe  permeance 
is  usually  estimated  graphically  by  drawing  lines  of  force,  taking 
Fig.  24  as  a  guide1 ;  the  permeance  of  each  tube  of  flux  between  the 
pole  and  the  armature  is.  fiA/l,  where  A  is  the  mean  cross-section, 
and  I  is  the  mean  length  of  the  tube.  The  fringe  permeance  is  of 
the  order  of  magnitude  of  10  per  cent  of  the  total  permeance  of 
the  air-gap,  so  that  some  error  in  its  estimation  does  not  seriously 
affect  the  total  required  ampere-turns.  Careful  designers  some- 
times calculate  the  air-gap  permeance  for  two  positions  of  the 
pole,  differing  from  each  other  by  one-half  of  the  tooth  pitch,  and 
take  the  average  of  the  two  results. 

Carter's  curve  could  be  used  directly  for  calculating  the  pole- 
fringe  permeance,  if  the  pole  waist  were  of  the  same  width  as 
the  pole  shoe  (line  mm'  in  Fig.  24),  and  if  the  armature  had  no 
slots.  In  this  case  the  space  between  the  adjacent  poles  could  be 
considered  as  a  big  slot,  and  the  curve  in  Fig.  26  could  be  directly 
applied  to  it.  On  account  of  a  smaller  width  of  the  pole  core  and 
because  of  the  armature  slots  the  mean  length  of  the  lines  of  force 
in  the  fringe  is  increased,  so  that  the  actual  permeance  of  the  pole- 
fringe  is  somewhat  smaller  than  that  according  to  Carter's  curve. 
By  practice  and  experience  one  can  acquire  a  judgment  as  to  what 
fraction  of  Carter's  permeance  to  take  in  a  given  case. 

The  length  le^  is  a  sum  of  the  parts  such  as  l\,  1'2,  etc.  (Fig.  24), 
on  which  the  lines  of  force  are  parallel,  and  of  small  additional 
lengths  which  take  account  of  the  fringing  in  the  air-ducts  and  at 
the  pole  flanks.  These  additional  lengths  are  again  estimated 
from  Carter's  curve  (Fig.  26).  The  fringe  Ad  in  an  air-duct  of 
the  width  d  is  practically  the  same  as  that  in  a  slot  of  the  width 
s  =  d.  The  additional  length  Af  for  the  pole  flanks  is  found  by  con- 
sidering the  two  fringes  as  due  to  a  slot  of  the  width  /.  When  the 
stationary  and  the  revolving  parts  are  of  the  same  axial  length  so 
that/=0,  there  still  remains  some  fringe  permeance  between  the 

1  See  Art.  41  below  in  regard  to  the  drawing  of  the  lines  of  force  by  the 
judgment  of  the  eye. 


CHAP.  V] 


EXCITING  AMPERE-TURNS 


95 


96  THE   MAGNETIC  CIRCUIT  [ART.  37 

flank  surfaces  of  the  two  iron  structures.  This  permeance  is,  how- 
ever, very  small,  and  has  to  be  estimated  empirically,  if  at  all. 

Strictly  speaking,  leff  is  different  for  each  tooth,  if  the  air-gap 
is  variable,  because  the  amount  of  fringing  in  the  air-ducts  and  at 
the  flanks  is  different.  However,  it  is  hardly  worth  the  effort 
in  ordinary  cases  to  calculate  leff  for  each  tooth.  It  is  sufficient  to 
take  an  average  leff  for  some  intermediate  value  of  the  air-gap. 

In  some  high-speed  alternators,  and  usually  in  induction 
motors,  air-ducts  are  provided  in  both  the  stationary  and  the 
revolving  parts,  in  the  same  planes.  The  flux  fringe  in  an  air-duct 
is  then  of  such  a  shape  that  the  lines  of  force  are  parallel  to  one 
another  in  the  middle  of  the  air-gap,  between  the  stator  and  the 
rotor.  Therefore,  when  using  the  curve  in  Fig.  26  for  such  a  case, 
the  cylindrical  surface  midway  between  the  stator  and  the  rotor 
must  be  taken  to  correspond  to  that  of  the  solid  iron  surface 
assumed  in  the  deduction  of  the  curve.  Hence,  \ax  must  be  used 
instead  of  ax  in  determining  Ai. 

Having  calculated  the  permeances  of  the  several  paths  per  pole 
pitch  the  total  permeance  of  the  air-gap  is  found  as  their  sum,  or 


(49) 


Then,  the  required  number  of  ampere-turns  is  determined  from 
eq.  (44).  The  method  gives  quite  correct  results,  especially  with 
some  experience  in  estimating  the  permeances  of  irregular  paths. 
Each  designer  usually  modifies  slightly  the  empirical  factors  which 
are  indispensable  in  this  method,  and  devises  short  cuts  good  for 
the  particular  kind  of  machine  in  which  he  is  interested. 

Instead  of  calculating  the  permeance  of  each  tooth  separately, 
some  engineers  replace  the  actual  variable  air-gap  by  an  equiva- 
lent constant  air-gap  aeg,  either  by  the  judgment  of  the  eye,  or  as 
in  prob.  18  above.  The  actual  peripheral  length  of  the  pole  arc  is 
increased  by  from  one  to  one  and  one-half  aeq  on  each  side  to  take 
into  account  the  fringing  at  the  pole-tips.  This  gives  the  number 
of  teeth  under  the  pole.  The  permeance  of  each  tooth  is  calcu- 
lated from  eq.  (48)  for  ax  =  aeq,  and  is  then  multiplied  by  the  num- 
ber of  teeth.  With  some  practice,  one  can  obtain  in  this  manner 
quite  accurate  results  at  a  considerable  saving  in  time. 

The  method  outlined  above  is  not  directly  applicable  to  induc- 
tion machines  which  have  slotted  cores  on  both  sides  of  the  air- 


CHAP.  V]  EXCITING  AMPERE-TURNS  97 

gap  (Fig.  23)  :  At  each  instant  some  stator  teeth  are  opposite  rotor 
teeth,  others  bridge  over  some  rotor  slots,  and  vice  versa.  The 
amount  of  overlap  varies  from  instant  to  instant,  causing  periodic 
fluctuations  in  the  air-gap  reluctance. 

Assume  first  that  both  the  stator  and  the  rotor  have  smooth 
surfaces  facing  the  air-gap.  Let  the  permeance  of  such  a  machine 
be  (P8.  If  now  the  armature  be  slotted,  the  cross-section  of 
the  paths  in  the  air-gap  (neglecting  the  fringe)  is  reduced  in  the 
ratio  ti/Ai  where  ti  and  AX  are  the  stator  tooth  width  and  tooth 
pitch  respectively.  The  permeance  (Ps  is  also  reduced  in  the  same 
ratio.  Let  the  rotor  be  also  provided  with  slots;  the  average 
cross-section  of  the  path  is  thereby  further  reduced  in  the  ratio 
(£2/^2),  where  t%  and  \2,  are  the  tooth  width  and  the  tooth  pitch  on 
the  surface  of  the  rotor.  Thus,  disregarding  the  spread  of  the 
flux,  the  average  air-gap  permeance  of  an  induction  motor  is 


the  symbol  (Pa  being  put  in  parentheses  to  indicate  that  a  further 
correction  for  the  tooth  fringe  is  necessary.1 

In  order  to  take  the  fringe  into  consideration,  an  empirical  cor- 
rection is  made  in  this  formula.  Namely,  it  is  assumed  that  the 
actual  permeance  of  the  fringes  of  the  stator  teeth  is  the  same  as 
if  the  rotor  had  a  smooth  core,  and  vice  versa.  Accordingly,  in  the 
preceding  formula,  the  values  ti  and  t%  of  the  tooth  widths  are 
corrected  for  the  fringe,  using  Carter's  curve  (Fig.  26).  The 
formula  becomes  then 

<Pa=«l'A)(«2'/A2)tf>,,        .....       (50) 

where  £/  and  tj  are  the  corrected  widths  of  the  stator  and  rotor 
teeth  respectively.  This  formula  has  been  found  to  be  in  a  satis- 
factory agreement  with  experimental  results.2 

1  For  a  more  rigorous  proof  of  this  formula  see  C.  A.  Adams,  "A  Study 
in  the  Design  of  Induction  Motors,"  Trans.  Amer.   Inst.  Electr.  Engs.,  Vol.  24 
(1905),  p.  335. 

2  T.  -F.  Wall,  The  Reluctance  of  the  Air-gap  in  Dynamo-machines,  Journ. 
Inst.  Electr.  Engrs.   (British),  Vol.  40  (1907-8),  p.  568.     E.  Arnold  in  his 
Wechselstromtechnik,  Vol.  5,  Part  1,  pp.  42,  43,  calculates  the  value  of  ka 
for  an  induction  machine  in  a  somewhat  different  way.     With  open  slots 
in  the  stator,  Arnold's  method  gives  lower  values  of  ka  than  they  are  in  reality. 
See  Hoock  and  Hellmund,  Beitrag  zur  Berechnung  des  Magnetizierungs- 


98  THE  MAGNETIC  CIRCUIT  [ART.  37 

Referring  to  Art.  36,  eq.  (50)  may  be  interpreted  as  follows: 
Let  ka'  be  the  air-gap  factor  for  the  slotted  stator  and  a  smooth- 
body  rotor;  let  ka"  be  the  same  factor  for  the  slotted  rotor  and  a 
smooth-body  stator.  Then  the  air-gap  factor  of  the  actual 
machine 

ka=ka'Xka".     .     .'   .....     (51) 

In  an  induction  motor  the  magnetic  flux  is  distributed  in  the 
air-gap  approximately  according  to  the  sine  law,  due  to  the  dis- 
tributed polyphase  windings.  Therefore  the  value  of  M  deter- 
mined from  eq.  (44)  gives  only  the  average  value  of  the  m.m.f. 
required  for  the  air-gap.  With  a  sine-wave  distribution  of  the 
flux  the  maximum  m.m.f.  is  x/2  times  larger  than  the  average 
value. 

Prob.  20.  What  is  the  permeance  of  the  air-gap  of  a  16-pole  direct- 
current  machine,  the  armature  of  which  has  a  diameter  of  250  cm.  and 
is  provided  with  324  slots,  1 2  by  1 5  mm .  ?  The  gross  length  of  the  armature 
is  23  cm.,  and  it  is  provided  with  three  ventilating  ducts,  10  mm.  wide 
each.  The  axial  length  of  the  poles  is  21.5  cm.  The  pole  shoes  cover 
05  per  cent  of  the  periphery,  and  are  not  chamfered.  The  length  of  the 
air-gap  is  10  mm.  Ans.  About  900  perm. 

Prob.  21.  The  machine  mentioned  in  problem  14  has  120  slots,  3  by 
6.5  cm.  The  pole-shoes  are  shaped  according  to  the  arc  of  a  circle  of  a 
radius  equal  to  90  cm.  and  subtending  36  degrees;  the  pole-tips  are  formed 
by  quadrants  of  a  radius  equal  to  2.5  cm.  Check  the  value  of  the  field 
current  (52  amp.)  given-  in  problem  15,  by  the  method  of  equivalent 
permeances. 

Prob.  22.  What  is  the  maximum  m.m.f.  across  the  air-gap  of  an 
induction  motor,  if  the  gross  average  flux  density  in  the  air-gap  (total 
flux. divided  by  the  gross  area  of  the  air-gap,  not  including  the  vents) 
is  3  kl./sq.cm.,  and  the  clearance  is  1.2  mm.?  The  bore  is  64  cm.;  the 
stator  is  provided  with  48  open  slots,  22  by  43  mm.  The  rotor  has  91 
half-closed  slots,  the  slot  opening  being  3  mm.  The  machine  has  a  vent 
7  mm.  wide  for  every  9  cm.  of  the  laminations. 

Ans.     820  amp.  turns. 

Prob.  23.  Show  that  At/a  =  1.2  +2.93  log  (s/2a),  if  the  fringing 
lines  of  force  are  assumed  to  be  concentric  quadrants  (Fig.  27,  to  the  left) 
with  the  points  c  as  the  center;  the  average  length  of  path  in  the  part 
6cc'  is  estimated  to  be  equal  to  1.2a,  and  the  average  width  0.72a.  Hint: 
The  permeance  of  an  infinitesimal  tube  of  force  of  a  radius  x  and  of  a 
width  dx  is  iL-dx/(%nx).  Integrate  this  expression  between  the  limits 

stromes  in  Induktionsmotoren,  Elektrotechnik  und  Maschinenbau,  Vol.  28 
(1910),  p.  743. 


CHAP.  V] 


EXCITING  AMPERE-TURNS 


99 


of  a  and  }s.     See  Adams,  loc.  cit.,  p.  332.    The  formula  can  be  used  only 
when  s  is  larger  than  2a. 

Prob.  24.     Show  that  J*/a  =  2.93  log  (1  +\*s/a),  if  the  fringing  lines 
of  force  are  assumed    to  consist    of  concentric    quadrants  (Fig.  27,  to 


c'        ]f 

a 

FIG.  27.— Two  simplified  paths  for  the  fringing  flux. 

the  right),    with  the  point  cr  as  a  center,    continued   as  straight  lines. 
See  Arnold,  Die  Gleichstrommachine,  Vol.  1,  p.  269. 

Prob.  25.  Show  that  formula  (50)  applies  to  synchronous  and  direct- 
current  machines  with  salient  poles  as  well,  if  tf  is  the  width  of  the 
pole  shoe,  corrected  for  the  fringe,  and  A2  is  the  pole  pitch. 


CHAPTER  VI 

EXCITING   AMPERE-TURNS   IN   ELECTRICAL 
MACHINERY—  (Continued) 

38.  The  Ampere-turns  Required  for  Saturated  Teeth.  The  teeth 
and  the  slots  of  an  armature,  under  the  poles,  are  magnetically  in 
parallel  (Fig.  24)  ;  hence,  part  of  the  flux  passes  from  the  pole  into 
the  armature  core  through  the  slots  between  the  teeth.  But,  with 
a  moderate  saturation  in  the  teeth,  say  below  18  kilolines  per 
square  centimeter,  the  amount  of  the  flux  which  passes  through  the 
slots  is  altogether  negligible.  If  the  taper  of  the  teeth  is  slight, 
the  required  ampere-turns  are  found  for  the  average  flux  density 
in  the  tooth,  taking  the  value  of  H  from  the  curves  in  Fig.  3. 

Should  the  taper  of  the  teeth  be  considerable,  as  is  the  case  in 
revolving  armatures  of  small  diameter,  the  flux  density  should  be 
determined  in  say  three  places  along  the  tooth,  viz.,  at  the  root, 
in  the  middle  part,  and  at  the  crown.  Let  the  corresponding 
values  of  magnetic  intensity  from  the  magnetization  curve  of  the 
material  be  H0,  Hm,  and  HI  .  Assuming  H  to  vary  along  the  tooth 
according  to  a  parabolic  law,  we  have,  according  to  Simpson's 
rule  in  the  first  approximation,  that  the  average  intensity  over  the 
tooth  is 

(52) 


If  a  greater  accuracy  is  desired,  the  values  of  H  can  be  determined 
for  more  than  three  cross-sections  of  the  tooth  and  Simpson's 
rule  applied.1  For  instance,  let  the  length  be  divided  into  n  equal 

1  A  designer  who  has  to  calculate  ampere-turns  for  teeth  frequently 
will  save  time  by  plotting  curves  for  the  average  H  against  the  flux  density 
BQ  at  the  root  of  the  teeth.  Each  curve  would  be  for  one  taper,  and  these 
curves  would  cover  the  usual  range  of  taper  in  the  teeth.  See  A.  Miller  Gray 
"Magnetomotive  Force  in  Non-uniform  Magnetic  Paths,"  Electrical  World, 
Vol.  57  (1911),  p.  111. 

100 


CHAP.  VI]  EXCITING  AMPERE-TURNS  101 

parts,  where  n  is  an  even  number.     Then,  we  have  that 


n-2)].     •     (53) 

When  the  flux  density  in  the  teeth  is  considerable,  say  between 
18  and  24  kilomaxwells  per  square  centimeter,  an  appreciable  part 
of  the  total  flux  passes  through  the  slots  between  the  teeth,  also 
through  the  air-ducts,  and  in  the  insulation  between  the  lamina- 
tions. Dividing,  therefore,  the  flux  per  tooth  pitch  by  the  net 
cross-section  of  the  tooth,  one  gets  only  the  so-called  apparent  flux 
density  in  the  tooth,  which  density  is  higher  than  the  true  density. 
With  highly  saturated  teeth,  a  small  difference  in  the  estimated 
flux  density  makes  an  appreciable  difference  in  the  required  number 
of  ampere-turns;  it  is  therefore  of  importance  to  know  how  to 
determine  the  true  density  in  a  tooth,  knowing  the  apparent 
density. 

Consider  first  the  case  of  a  machine  with  a  large  diameter,  in 
which  the  taper  of  the  teeth  can  be  neglected.  Assume  the  con- 
centric cylindrical  surfaces  at  the  tips  and  at  the  roots  of  the  teeth 
to  be  equipotential  surfaces,  and  the  lines  of  force  to  be  all  parallel 
to  each  other,  in  the  slots  as  well  as  in  the  iron.  In  reality,  some 
lines  of  force  enter  the  teeth  on  the  sides  of  the  slots  (Fig.  24),  so 
that  the  foregoing  assumptions  are  not  quite  correct  ;  but  they  are 
the  simplest  ones  that  can  be  made.  Any  other  assumptions 
would  lead  to  calculations  too  complicated  for  practical  use. 

Let  Breal  be  the  true  flux  density  in  the  iron  of  the  tooth,  and 
let  Bapp  be  the  apparent  flux  density  in  the  tooth  under  the  assump- 
tion that  no  flux  passes  through  the  slots,  air-ducts,  or  insulation 
between  the  laminations.  Then,  denoting  the  actual  flux  density 
in  the  air  by  Ba,  we  have  the  following  expression  for  the  total 
flux  per  tooth  pitch: 


==  A  ^n 


reai 


where  Ai  and  Aa  are  the  cross-sections  in  square  centimeters  of  the 
paths  per  tooth  pitch,  in  the  iron  and  air  respectively.  Since  the 
iron  and  the  air  paths  are  of  equal  length,  and  are  in  parallel,  the 
m.m.f.  gradient  is  the  same  in  both.  Let  H  be  this  gradient,  in 


102  THE   MAGNETIC   CIRCUIT     .  [ART.  38 

kiloampere-turns  per  centimeter.      Then,   if  all  the  flux  densities 
are  in  kilomaxwells  per  square  centimeter,. 


Substituting  this  value  of  Ba  into  the  preceding  equation  we 
obtain,  after  division  by  At-, 

BapP  =  Breal-^\^(Aa/A^H  .....     (54) 

The  ratio  Aa/Ai  can  be  expressed  through  the  dimensions  of 
the  machine  as  follows:  Ai^tln  where  t  is  the  width  of  the  tooth, 
and  ln  is  the  net  axial  length  of  the  laminations,  without  the  air- 
ducts  and  insulation.  Aa=XLg—tln,  where  X  is  the  tooth  pitch 
(Fig.  20),  and  lg  is  the  gross  length  of  the  armature  core.  Hence, 


(55) 


In  eq.  (54)  the  flux  density  Bapp  and  the  ratio  AJAi  are  known 
in  any  particular  case,  and  the  problem  is  to  find  Breal  and  H: 
The  other  equation  which  connects  Breal  and  H  is  the  magnetiza- 
tion curve  of  the  material,  and  the  problem  can  be  solved  in  a  simi- 
lar manner  to  problem  1  1  in  chapter  II  (see  also  problem  4  below)  . 

Professional  designers  use  curves  like  those  shown  in  Fig.  28, 
which  give  directly  the  relation  between  Bapp  and  Breai  within  the 
range  of  values  of  Aa/Ai  which  occur  in  practice.  The  curves  are 
plotted  point  by  point  by  assuming  certain  values  of  Breal  and  cal- 
culating the  corresponding  Bapp  from  eq.  (54).  For  instance,  for 
jBreaj=24,  the  saturation  curve  shown  in  Fig.  28  gives  H=  1.33,  so 
that  for  Aa/Ai=2,  we  have:  £opp  =  24  +  1.25X2X1.33  =  27.33. 
This  determines  one  point  on  the  curve  marked  "  Ratio  of  air  to 
iron  =  2."  In  using  these  curves  one  begins  with  the  known  value 
of  Bapp  on  the  lower  axis  of  abscissae,  and  follows  the  ordinate  to 
the  intersection  with  the  curve  for  the  desired  ratio  Aa/  '  AI\  this 
gives  the  value  of  Breal.  By  following  the  horizontal  line  from  the 
point  so  located  to  the  intersection  with  the  B-H  curve,  the  corre- 
sponding value  of  H  is  read  off  on  the  upper  axis  of  abscissae. 

The  curves  in  Fig.  28  are  completely  determined  by  the  shape 
of  the  B-H  curve,  so  that,  if  the  material  to  be  used  for  the  arma- 
ture core  differs  considerably  from  that  assumed  in  Fig.  28,  new 
curves  of  Breal  versus  Bapp  ought  to  be  plotted,  or  else  the  method 


CHAP.  VI] 


EXCITING  AMPERE-TURNS 


103 


jL 

\T\ 


\ 


\ 


\ 


\ 


\\ 


S 


3 


% 


g 

I 

S 

^r 

^H 


104  THE  MAGNETIC  CIRCUIT  [ART.  38 

may  be  used  which  is  suggested  in  problem  4  below.  A  comparison 
of  the  B-H  curve  with  those  in  Fig.  3  shows  that  a  much  better 
quality  of  steel  is  presupposed  in  Fig.  28.  Such  is  usually  the  case 
when  it  is  desired  to  employ  highly  saturated  teeth,  for  otherwise 
it  might  be  practically  impossible  to  get  the  required  flux.  The 
curves  in  Fig.  3  refer  to  an  average  quality  of  electrical  steel. 

Formula  (54)  and  the  curves  in  Fig.  28  presuppose  that  the 
teeth  have  no  taper,  or  that  the  taper  is  negligible.  If  the  taper  of 
the  teeth  is  quite  considerable  the  tooth  and  the  slot  are  divided 
by  equipotential  cylindrical  surfaces  into  two  or  more  parts,  and 
H  is  determined  separately  for  each  part.  Then  the  effective  value 
of  H  is  calculated  according  to  Simpson's  rule,  using  either  formula 
(52)  or  (53). 

Prob.  1.  A  four-pole  direct-current  armature  has  the  following 
dimensions :  diameter  45  cm. ;  gross  length  of  core  20  cm. ;  two  air-ducts 
7  mm.  each;  67  open  slots  1  by  3  cm.  The  poles  are  of  such  a  shape 
that  the  flux  per  pole  is  carried  uniformly  by  11.5  teeth.  How  many 
ampere-turns  per  pole  are  required  for  the  teeth  when  the  flux  per  pole 
is  3  megalines?  Use  the  saturation  curve  for  carbon-steel  laminations 
in  Fig.  3.  Ans.  148  amp  .-turns. 

Prob.  2.  How  many  ampere-turns  are  required  in  the  preceding 
problem  when  the  flux  per  pole  is  4.4  megalines? 

Ans.  Between  2400  and  2500. 

Prob.  3.  The  machine,  in  problem  22  of  the  preceding  chapter,  had 
a  gross  average  flux  density  in  the  air-gap  of  3  kl./sq.  cm.  The  bore 
was  64  cm.  The  stator  was  provided  with  48  slots  22  by  43  mm.  The 
machine  has  a  vent  7  mm.  wide  for  every  9  cm.  of  the  laminations. 
What  is  the  maximum  m.m.f .  required  for  the  stator  and  rotor  teeth,  if 
the  size  of  each  of  the  91  rotor  slots  is  14  by  30  mm.  below  the  overhang? 
Ans.  between  2400  and  2500  amp. -turns. 

Prob.  4.  Instead  of  drawing  the  curves  shown  in  Fig.  28,  the  relation 
between  Breaz  and  Bapp  can  be  found  by  the  following  construction: 
Disregard  the  lower  scale  marked  "  Apparent  Flux  Density  ";  extend 
the  left-hand  scale  to  the  division  34  and  mark  the  scale  "  Real  and  Appar- 
ent flux  density."  Cut  out  a  strip  of  paper,  and  copy  the  left-hand  scale 
on  the  left-hand  edge  of  the  strip.  On  the  right-hand  edge  of  the  strip 
mark  the  scale  for  Aa/Ai  as  follows:  division  26  of  the  flux  density  to 
correspond  with  zero,  division  27  with  0.4,  division  28  with  0.8,  etc. 
Apply  the  left-hand  edge  of  the  strip  to  division  2.0  on  the  upper  horizon- 
tal scale,  and  to  division  26  on  the  lower  horizontal  scale.  Move  the 
strip  up  and  down  until  the  upper  horizontal  scale  coincides  with  the 
desired  value  of  Aa/At  marked  on  the  strip.  Lay  a  straightedge  on  the 
divisions  of  the  two  vertical  scales  corresponding  to  the  given  apparent 
flux  density.  The  intersection  of  the  straightedge  with  the  B-H  curve 


CHAP.  VI]  EXCITING  AMPERE-TURNS  105 

will  give  the  required  values  of  Breai  and  H.  Check  this  construction 
for  a  few  points  with  the  values  obtained  from  the  curves,  and  give  a 
general  proof.  Hint:  This  construction  amounts  to  considering  the 
B-H  curve  and  Eq.  (54)  as  two  simultaneous  equations  with  two  unknown 
quantities  Breai  and  H.  See  problem  11  in  chapter  II,  Art.  13. 

39.  The  Ampere-turns  for  the  Armature  Core  and  for  the 
Field  Frame.  In  many  machines  the  m.m.f.  required  for  the  air- 
gap  and  the  teeth  are  large  as  compared  to  those  required  for  the 
armature  core  and  the  field  frame;  in  such  cases  the  latter  are 
either  altogether  neglected,  or  are  estimated  roughly,  by  increas- 
ing the  ampere-turns  calculated  for  the  rest  of  the  magnetic  cir- 
cuit by  say  five  or  ten  per  cent.  Where  this  is  not  permissible,  the 
usual  procedure  is  to  estimate  the  maximum  flux  density  in  the 
core  or  frame  under  consideration  and  to  measure  from  the  drawing 
of  the  machine  the  length  of  the  average  path  of  the  lines  of  force 
in  it.  The  assumption  is  made  that  the  same  flux  density  is  main- 
tained on  the  whole  length  of  the  path,  and  the  required  ampere- 
turns  are  calculated  from  the  magnetization  curve  of  the  material 
(Figs.  2  and  3).  While  the  ampere-turns  determined  in  this  way 
are  usually  larger  than  those  actually  required,  the  method  is  per- 
missible if  the  total  amount  of  the  m.m.f.  for  the  parts  under  con- 
sideration is  small  as  compared  to  the  total  m.m.f.  of  the  magnetic 
circuit.  If  a  greater  accuracy  is  desired,  the  path  is  subdivided 
into  two  or  more  parts  in  series,  and  the  average  density  deter- 
mined for  each  part;  and  then  the  ampere-turns  required  for  each 
part  are  added. 

The  tendency  now  is  to  increase  the  flux  density  in  the  arma- 
ture cores  of  alternators  and  induction  motors  so  as  to  reduce  the 
size  of  the  machine.  This  is  made  possible  through  a  better  quality 
of  laminations,  which  show  a  smaller  core  loss,  and  also  through  the 
use  of  a  more  intensive  ventilation.  With  these  high  densities 
and  with  the  comparative  large  values  of  the  pole  pitch  necessary  in 
high-speed  machinery,  the  ampere-turns  for  the  core  constitute  an 
appreciable  amount  of  the  total  m.m.f.  of  the  machine,  and  it  is 
therefore  desirable  to  calculate  them  more  accurately. 

The  flux  density  in  the  core  is  a  minimum  opposite  the  center 
of  a  pole,  and  is  a  maximum  in  the  radial  plane  midway  between 
two  poles  (Fig.  15) .  At  each  point  the  flux  density  has  a  tangen- 
tial and  a  radial  component.  The  latter  is  comparatively  small 
and  can  be  neglected ;  the  tangential  component  can  be  assumed 


106  THE   MAGNETIC   CIRCUIT  [ART.  39 

to  vary  according  to  the  sine  law,  being  zero  opposite  the  center  of 
the  pole  and  reaching  its  maximum  between  the  poles.  With 
these  assumptions,  knowing  the  maximum  flux  density  in  the  core, 
the  flux  density  at  all  other  points  is  calculated,  and  the  corre- 
sponding values  of  H  are  determined  from  the  B-H  curve  of  the 
material.  The  average  value  of  H  for  one-half  pole  pitch  is  then 
found  by  Simpson's  rule,  eqs.  (52)  and  (53).  With  the  sine-wave 
assumption,  the  average  H  depends  only  upon  the  maximum 
flux  density,  so  that  for  a  given  material  a  curve  can  be  compiled 
from  the  B-H  curve,  giving  directly  Have  for  different  values  of 
/?  i 

•L-'max' 

Should  a  still  greater  accuracy  be  required,  the  following 
method  can  be  used:  Draw  the  assumed  or  the  calculated  curve 
of  the  distribution  of  flux  density  in  the  air-gap,  and  indicate  to 
your  best  judgment  the  tubes  of  force  in  the  armature  core,  say 
for  each  tooth  pitch.  The  flux  in  the  radial  plane  midway  between 
the  two  poles  can  be  assumed  to  be  distributed  uniformly  over  the 
cross-section,  and  this  fact  facilitates  greatly  the  determination  of 
the  shape  of  the  tubes  of  flux.  The  m.m.f .  required  for  each  tube 
is  calculated  by  dividing  it  into  smaller  tubes  in  series  and  in  paral- 
lel; thus,  either  the  average  m.m.f.  for  the  whole  flux  can  be  found, 
or  the  maximum  m.m.f.  for  one  particular  tube.2 

The  frame  to  which  the  poles  are  fastened  in  direct-current  and 
in  synchronous  machines  is  usually  made  of  cast  iron;  in  some 
cases  the  frame  is  made  of  cast  steel;  in  high-speed  synchronous 
machines  the  revolving  field  is  made  of  forged  steel.  The  magneto- 
motive force  required  for  such  a  frame  is  found  in  the  usual  way 
from  the  magnetization  curve  of  the  material,  knowing  the  area 
and  the  average  length  of  the  path  between  two  poles;  the  length 
is  estimated  from  the  drawing  of  the  machine.  In  figuring  out  the 
flux  density  in  a  field  frame  one  must  not  forget  that  (1)  only  one- 
half  of  the  flux  per  pole  passes  through  a  given  cross-section  of  the 
frame  (Fig.  20) ;  (2)  the  total  flux  in  the  frame  and  in  the  poles  is 
larger  than  that  in  the  armature  by  the  amount  of  the  leakage  flux 
between  the  poles.  This  leakage  is  usually  estimated  in  per  cent 

1  This  method  is  due  to  E.  Arnold.     See  his  Wechselstromtechnik,  Vol.  5, 
(1909),  part  1,  p.  48. 

2  For  details  of  this  method  see  Hoock  and  Hellmund,  Beitrag  zur  Berech- 
nung    des  Magnetiziemngsstromes  in  Induktionsmotoren,  Elektrotechnik  und 
Maschinenbau,  Vol.  28  (1910),  p.  743. 


CHAP.  VI]  EXCITING  AMPERE-TURNS  107 

of  the  useful  flux,  from  one's  experience  with  previously  built 
machines,  or  it  can  be  calculated  by  the  methods  explained  in  the 
next  article.  Thus,  a  leakage  factor  of  1.20  means  that  the  flux  in 
the  field  poles  is  20  per  cent  higher  than  that  in  the  armature,  the 
leakage  flux  constituting  20  per  cent  of  the  useful  flux.  The  usual 
values  of  the  leakage  factor  vary  between  1.10  and  1.25,  depending 
upon  the  proximity  of  the  adjacent  poles,  the  degree  of  saturation 
of  the  circuit,  and  the  proportions  of  the  machine. 

The  ampere-turns  required  for  the  pole-pieces  are  calculated  in 
a  similar  way,  assuming  the  whole  leakage  to  take  place  between 
the  pole-tips,  so  that  the  flux  density  in  the  pole-waist  corresponds 
to  the  total  flux;  including  the  leakage  flux.  In  exceptional  cases 
of  highly  saturated  pole-cores  this  method  may  be  inadmissible,  on 
account  of  too  large  a  margin  which  it  would  give  as  compared  to 
the  ampere-turns  actually  required.  In  such  cases  part  of  the 
leakage  may  be  assumed  to  be  concentrated  between  some  two 
corresponding  points  on  the  waists  of  two  adjacent  poles,  or  it  may 
be  assumed  to  be  actually  distributed  between  the  two  pole-waists. 
See  probs.  9  and  10  in  chapter  II. 

In  some  machines  the  joint  between  the  pole  and  the  frame 
offers  a  perceptible  reluctance,  like  the  joints  in  the  transformer 
cores  discussed  in  Art.  33.  Some  designers  allow  a  certain 
fraction  of  a  millimeter  of  air-gap  to  account  for  this  reluctance, 
and  add  the  number  of  ampere-turns  required  to  maintain  the 
flux  in  this  air-gap  to  those  for  the  pole-piece.  The  length  of 
this  equivalent  air-gap  is  found  by  checking  back  no-load 
saturation  curves  obtained  from  experiment.  As  a  usual  rule, 
it  is  advisable  to  increase  the  total  calculated  ampere-turns  of  the 
magnetic  circuit  by  about  5  to  10  per  cent.  This  increase  covers 
such  minor  points  as  the  reluctance  of  the  joints,  omitted  .in 
calculations,  as  well  as  certain  inaccurate  assumptions;  it  also 
covers  a  possible  discrepancy  between  the  assumed  and  the 
actual  permeability  of  the  iron.  With  a  liberally  proportioned 
field  winding  and  a  proper  regulating  rheostat  a  designer  can 
rest  assured  that  the  required  voltage  will  be  obtained,  though 
possibly  at  a  somewhat  different  value  of  the  field  current  than 
the  estimated  one. 

Prob.  5.  The  stator  core  of  a  six-pole  induction  motor  has  the 
following  dimensions:  bore  112  cm.;  outside  diameter  145  cm.;  gross 
length  55  cm. ;  the  slots  are  2  cm.  X4.5  cm. ;  the  machine  is  provided  with 


108  THE  MAGNETIC   CIRCUIT  [ART.  40 

8  ventilating  ducts  9  mm.  wide  each.    What  is  the  maximum  m.m.f. 
required  for  the  stator  core  per  pole  if  the  flux  per  pole  is  0.15  weber? 

Ans.     190  using  Arnold's  method. 

Prob.  6.  Draw  a  curve  between  the  average  H  and  maximum  B 
in  the  core,  assuming  a  sinusoidal  distribution  of  the  flux  density  in  the 
tangential  direction,  for  the  carbon  steel  laminations  in  Fig.  3. 

Ans.     Have  =  26.5  for  Bmax  =  18. 

Prob.  7.  The  cross-section  of  the  cast-iron  field  yoke  of  a  direct- 
current  machine  is  370  sq.cm.;  the  mean  length  of  path  in  it  between 
two  consecutive  poles  is  85  cm.  The  length  of  the  lines  of  force  in  each 
pole-waist  is  21  cm.;  its  cross-section  420  sq.cm.  The  poles  are  made 
of  steel  laminations  4  mm.  thick,  so  that  the  space  lost  between  the 
laminations  is  negligible.  The  reluctance  of  the  joint  between  a  bolted 
pole  and  the  yoke  is  estimated  to  be  equivalent  to  0.1  mm.  of  air.  What 
is  the  required  number  of  ampere-turns  for  the  pole-piece  and  the  yoke, 
per  pole,  when  the  useful  flux  of  the  machine  is  5  megalines  per  pole? 
The  leakage  factor  is  estimated  to  be  equal  to  1.20. 

Ans.    About  930. 

40.  Magnetic  Leakage  between  Field  Poles.  It  is  of  impor- 
tance in  modern  highly  saturated  machines  to  know  accurately  the 
leakage  flux  between  the  poles,  in  order  to  estimate  correctly  the 
ampere-turns  required  for  the  field  poles  and  the  frame  of  the 
machine.  Moreover,  the  design  of  the  poles  can  be  improved, 
knowing  exactly  where  the  principal  leakage  occurs  and  how  it 
depends  upon  the  proportions  of  the  machine.  The  value  of  the 
leakage  factor  also  affects  the  voltage  regulation  of  the  machine, 
because  at  full  load  the  m.m.f.  between  the  pole-tips  has  to  be 
larger  than  at  no-load,  on  account  of  the  armature  reaction. 

For  new  machines  of  usual  proportions  the  value  of  the  leakage 
factor  can  be  estimated  from  tests  made  upon  similar  machines. 
But  in  new  machines  of  unusual  proportions  the  designer  has  to 
rely  upon  his  judgment,  assisted  if  necessary  by  crude  compara- 
tive computations  of  the  permeance  between  adjacent  poles.  In 
this  and  in  the  next  article  some  examples  of  such  computations 
are  given,  not  so  much  in  order  to  give  a  definite  method  to  be  fol- 
lowed in  all  cases,  as  to  show  the  student  a  possible  procedure  and 
to  train  his  judgment  in  estimating  the  permeance  of  an  irregular 
path. 

Four  principal  paths  of  leakage  can  be  distinguished  between 
two  adjacent  poles  (Fig.  29) :  (a)  between  the  sides  of  the  pole 
shoes  which  face  each  other;  (6)  between  the  sides  of  the  pole 
cores  (waists)  parallel  to  the  shaft  of  the  machine;  (c)  between  the 


CHAP.  VI] 


EXCITING  AMPERE-TURNS 


109 


flanks  or  sides  of  the  pole  shoes  perpendicular  to  the  shaft;  (d) 
between  the  flanks  of  the  pole  cores.  In  the  calculations  which 
follow,  the  permeances  are  computed  between  a  pole  and  the  planes 
of  symmetry,  MN,  between  the  two  poles,  the  permeance  of  the 
other  half  of  each  path  being  the  same.  All  these  leakage  paths 
are  in  parallel  with  respect  to  the  pole,  so  that  the  total  leakage 


(c) 


FIG.  29.— The  leakage  flux  between  field  poles. 

permeance  is  equal  to  their  sum.  Knowing  this  total  permeance 
and  the  m.m.f.  between  the  pole  and  the  plane  MN  the  leakage 
flux  is  found,  and  knowing  this  flux  and  the  useful  flux  per  pole 
the  leakage  factor  is  easily  calculated. 

We  shall  now  estimate  the  permeances  of  each  of  the  four  above 
mentioned  paths  of  leakage. 

(a)  Between  the  adjacent  pole-tips.  Estimate  the  average 
cross-section  A  of  the  path,  in  square  centimeters,  and  the  aver- 


110  THE  MAGNETIC  CIRCUIT  [ART.  40 

age  length  I  between  the  pole-tip  and  the  plane  MN,  being  guided 
by  Fig.  29.  (Do  not  encroach  upon  the  fringe  to  the  armature.) 
Then  the  permeance  of  the  path  is,  in  perms, 


.     .     ....     .     (56) 

If  a  greater  accuracy  is  desired,  subdivide  the  total  path  into 
smaller  paths  in  series  and  in  parallel,  and  calculate  the  permeance 
or  the  reluctance  of  each  separately.  Then  the  total  permeance  is 
found  according  to  the  well-known  law  of  combination  of  reluc- 
tances and  permeances  in  series  and  in  parallel  (Art.  9).  When 
mapping  out  the  lines  of  force  in  the  air,  begin  them  nearly  at  right 
angles  to  the  surface  of  the  pole  (see  Art.  41  a  below)  and  draw 
them  so  as  to  make  the  total  permeance  of  the  path  a  maximum, 
that  is,  reducing  as  far  as  possible  the  length  and  increasing  the 
cross-section  of  each  elementary  tube  of  flux.  The  medium  may 
be  said  to  be  in  a  state  of  tension  along  the  lines  of  force,  and  of 
compression  at  right  angles  to  their  direction,  by  virtue  of  the 
energy  stored  in  the  field.  Hence,  there  is  a  tendency  for  the  tubes 
of  force  to  contract  along  their  length  and  expand  across  their 
width. 

(b)  Between  the  opposite  pole-cores.     In  this  part  of  the  leakage 
field  each  elementary  concentric  path  is  subjected  to  a  different 
m.m.f.,  that  between  the  roots  of  the  poles  being  practically  zero, 
while  the  m.m.f.  between  the  points  p  and  p'  is  equal  to  that 
between  the  pole-tips.     In  most  cases  it  is  permissible  to  consider 
the  whole  leakage  flux  as  if  passing  through  the  whole  length  of  the 
pole  core,  and  then  crossing  to  the  adjacent  poles  at  the  pole-tips. 
Therefore,  it  is  convenient  to  add  the  permeance  between  the  pole 
cores  to  that  between  the  pole-tips.     But  the  average  m.m.f. 
between  the  waists  is  only  about  one-half  of  that  between  the  tips, 
so  that  the  equivalent  permeance  between  the  pole  cores,  reduced 
to  the  total  m.m.f.,   is  equal  to  one-half  of  the  actual  permeance. 
If  the  actual  permeance,  calculated  according  to  formula  (56)  is  (P, 
the  effective  permeance  is  \(P.     The  average  length  and  cross- 
section  of  the  path  are  easily  estimated  from  the  drawing  of  the 
machine. 

(c)  Between  the  flanks  of  the  pole  shoes.    The  path  extends  indefi- 
nitely outside  the  machine,  and  the  lines  of  force  are  twisted 
curves,  so  that  it  is  difficult  to  estimate  the  permeance  graphically. 
As  a  rough  estimate,  this  permeance  can  be  reduced  to  that  of  the 


CHAP.  VI] 


EXCITING  AMPERE-TURNS 


111 


path  in  the  air  between  two  rectangular  poles  of  an  electromagnet 
(Fig.  30).  Assume  the  paths  of  the  flux  to  consist  of  concentric 
quadrants  with  the  centers  at  c  and  c',  joined  by  parallel  straight 
lines,  and  let  the  width  of  the  poles  in  the  direction  perpendicular 
to  the  plane  of  the  paper  be  h.  Then  the  permeance  of  an  infin- 
itesimal layer  of  thickness  dx,  between  one  of  the  poles  and  the 
plane  MN  of  symmetry,  is 

d(P 


Integrating  this  expression  between  the  limits  o  and  b  we  find 

(P=  IMh  log  (1.57&/Z  +  1)  perms  .     .     .     .     (57) 

(compare  with  prob.  24  in  Chapter  V,  Art.  37). 

In  applying  this  formula  and  Fig.  30  to  the  case  of  the  flank 
leakage  between  the  pole  shoes,  h  is  the  average  radial  height  of 
the  pole  shoe,  6  is  equal  to  one- 
half  the  width  of  the  pole  shoe, 
and  21  is  the  distance  between 
the  two  opposing  pole-tips. 
While  the  method  evidently 
gives  only  a  crude  approxi- 
mation to  the  actual  perme- 
ance, formula  (57)  at  least 
fixes  a  lower  limit  to  the  per- 
meance in  question. 

(d)  Between  the  flanks  of 
the  pole  cores.  The  conditions 
are  similar  to  those  under  (c), 
so  that  the  permeance  is  esti- 
mated again  on  the  basis  of 
formula  (57)  .  The  sides  of  the 
two  rectangles  in  Fig.  29  are 
not  parallel  to  each  other  as  in 
Fig.  30,  but  this  difference  is 
taken  into  account  by  mentally 
turning^  them  into  a  parallel 

position,  and  estimating  the  equivalent  distance  21  between  the 
edges  of  the  opposing  poles.  The  dimension  h  is  in  this  case  the 
radial  height  of  the  pole-waist,  and  b  is  one-half  of  the  width  of 
the  pole-waist.  The  flank  leakage  is  smaller  than  that  between 


FIG.  30.— The  magnetic  path  between 
the  poles  of  an  electromagnet. 


112  THE  MAGNETIC  CIRCUIT  [ART.  40 

the  opposite  side,  so  that  one  may  be  satisfied  with  a  lesser  degree 
of  accuracy.  The  equivalent  permeance,  reduced  to  that  between 
the  pole-tips,  is  again  equal  to  one-half  the  actual  permeance,  for 
the  same  reason  as  under  (6)  above. 

The  total  leakage  permeance  between  a  pole  and  the  two  planes 
of  symmetry  is  equal  to  the  sum  of  the  permeances  calculated  as 
above.  In  summing  them  up  it  will  be  seen  from  Fig.  29  that  the 
permeances  (a)  and  (b)  must  be  taken  twice,  and  also  that  (c)  and 
(d)  must  be  taken  four  times.  The  leakage  flux  per  pole  is 
obtained  by  multiplying  the  total  leakage  permeance  by  the  m.m.f . 
between  the  pole-tip  and  the  plane  of  symmetry.  This  m.m.f.  is 
equal  to  that  required  to  establish  the  useful  flux,  along  the  path 
qrs,  through  the  air-gap  and  the  armature  of  the  machine,  and 
consequently  it  is  known  before  the  pole-piece  and  the  field  wind- 
ing are  computed  in  detail.  Knowing  the  leakage  flux  and  the 
useful  flux,  the  leakage  factor  is  figured  out  according  to  the  defi- 
nition given  above. 

When  calculating  permeances  as  indicated  above,  one  is  advised 
to  make  liberal  estimates  of  the  same,  for  two  reasons :  In  the  first 
place, the  true  permeance  of  a  path  is  always  the  largest  possible,  so 
that,  whatever  assumptions  one  makes,  the  calculated  permeance 
comes  out  smaller  than  the  actual.  In  the  second  place,  in  design- 
ing a  new  machine  it  is  better  to  be  on  the  safe  side  and  rather 
underestimate  than  overestimate  the  excellence  of  the  perform- 
ance. Some  writers  give  more  elaborate  rules  and  formulae  for  the 
calculation  of  the  leakage  permeance  which  are  useful  in  the 
design  of  machines  of  special  importance.1 

The  leakage  factor  remains  practically  constant  as  long  as  the 
flux  density  in  the  armature  core  and  teeth  is  moderate,  so  that 
the  reluctance  of  the  useful  path  qrs  is  nearly  constant.  This  is 
because  the  reluctance  of  the  leakage  paths  is  constant,  and,  if  the 
reluctance  of  the  useful  path  is  also  constant,  the  useful  flux  and 
the  leakage  flux  increase  in  the  same  proportion  when  the  m.m.f. 
between  the  pole-tips  is  increased.  When  the  armature  iron  is 
approaching  saturation,  the  leakage  factor  increases  with  the  field 

1  For  a  more  detailed  treatment  of  the  leakage  between  poles  see  the 
following  works:  E.  Arnold,  Die  Gleichstrommaschine,  Vol.  1  (1906),  pp. 
284-294;  Hawkins  and  Wallis,  The  Dynamo,  Vol.  1  (1909),  pp.  469-484; 
Pichelmayer,  Dynamobau  (19Q8),  pp.  127-131;  Cramp,  Continuous-Current 
Machine  Design  (1910),  pp.  42-47  and  226-230. 


CHAP.  VI]  EXCITING  AMPERE-TURNS  113 

current,  because  the  leakage  flux  increases  the  more  rapidly  than 
the  useful  flux.  This  increase  is  partly  offset  by  the  fact  that  the 
pole-tips  also  become  gradually  saturated  by  the  leakage  flux,  so 
that  the  leakage  factor  does  not  increase  as  rapidly  as  it  would 
otherwise.  The  practical  point  to  be  observed  is,  that  for  the 
higher  flux  densities,  if  accuracy  is  desired,  the  leakage  should 
be  estimated  separately  for  a  few  points  on  the  no-load  saturation 
curve. 

For  a  given  terminal  voltage,  the  leakage  factor  of  a  machine  is 
somewhat  higher  at  full-load  than  at  no-load,  because  the  required 
m.m.f .  between  the  pole-faces  is  higher,  due  to  the  armature  reac- 
tion and  to  the  voltage  drop  in  the  armature.  In  comparatively 
rare  cases,  when  the  armature  reaction  assists  the  field  m.m.f.,  for 
instance,  in  the  case  of  an  alternator  supplying  a  leading  current, 
the  leakage  factor  decreases  with  the  increasing  load.  The  fol- 
lowing example  illustrates  the  influence  of  the  load  upon  the  value 
of  the  leakage  factor. 

Let  the  useful  flux  per  pole  in  an  alternator,  at  the  rated 
voltage  and  at  no-load,  be  5  megalines,  and  let  6000  amp. -turns 
per  pole  be  required  for  the  air-gap  and  the  armature  core.  Let 
the  permeance  of  the  leakage  paths  between  a  pole  and  the  neutral 
planes  be  120  perms,  so  that  the  leakage  flux  is  0.72  megaline,  and 
the  leakage  factor  is  (5.00 +0.72)/5.00=  1.14.  Let  a  useful  flux 
of  5.5  megalines  be  required  at  the  same  voltage  and  at  full  load, 
an  increase  of  10  per  cent  being  necessary  to  compensate  for  the 
internal  drop  of  voltage  due  to  the  armature  impedance.  If  the 
teeth  and  the  armature  core  were  not  saturated  at  all,  an  m.m.f. 
of  6600  amp.-turns  would  be  required.  In  reality,  the  m.m.f.  is 
higher,  say  7500  amp.-turns.  Let  the  armature  reaction  be  equal 
to  1500  demagnetizing  ampere-turns  per  pole.  To  compensate 
for  its  action,  1500  additional  ampere-turns  are  required  on  each 
field  coil.  Thus,  the  difference  of  magnetic  potential  between 
a  pole-tip  and  the  adjacent  plane  of  symmetry  MN  (Fig.  29) 
is  now  9000  amp.-turns,  and  the  leakage  flux  is  increased  to 
1.08  megalines.  Therefore,  the  leakage  factor  at  full  load  is 
(5.50  + 1. 08) /5.50=  1.20.  Similar  relations  hold  for  the  direct- 
current-  machines. 

In  calculating  the  performance  of  a  synchronous  or  a  direct- 
current  machine  one  has  to  use  the  relation  between  the  field  cur- 
rent and  the  voltage  induced  in  the  armature.  Ordinarily,  the 


114  THE   MAGNETIC   CIRCUIT  [ART.  40 

no-load  saturation  curve  is  used  for  this  purpose,  assuming  that 
the  leakage  factor  is  the  same  at  full  load  as  at  no  load.  However 
careful  designers  sometimes  plot  a  separate  curve,  using  a  higher 
leakage  factor,  for  use  at  full  load. 

Prob.  8.  Assume  in  the  illustrative  example  given  in  the  text  the 
armature  current  to  be  leading,  so  that  the  voltage  drop  in  the  armature 
is  negative  and  the  armature  reaction  strengthens  the  field.  Show  that 
with  the  same  value  of  the  armature  current  the  leakage  factor  is  about 
1.09. 

Prob.  9.  Draw  rough  sketches  of  the  magnetic  circuits  of  two 
machines,  one  possessing  such  proportions,  number  and  shape  of  poles 
as  to  give  a  particularly  low  leakage  factor,  the  other  markedly  deficient 
in  this  respect. 

Prob.  10.  Calculate  the  leakage  factor  and  the  leakage  permeance 
per  pole  of  a  six-pole  turbo-alternator  of  the  following  dimensions;  the 
bore  is  1.2  m.;  the  axial  length  of  the  poles  0.6  m.;  minimum  air-gap 
1  cm.;  maximum  air-gap  2  cm.;  total  height  of  the  pole  23  cm.;  the 
height  of  the  pole-waist  18  cm.;  the  breadth  across  the  pole-waist  25  cm.; 
that  across  the  pole-tips  36  cm.  The  reluctance  of  the  useful  path  in 
the  air-gap  and  in  the  armature  is  estimated  to  be  about  0.57  millirel 
per  pole.  Ans.  1.115;  about  200  perms. 

Prob.  11.  The  leakage  factor  of  the  machine  specified  in  the  pre- 
ceding problem  was  found  from  an  experiment  to  be  1.13,  at  no-load, 
when  the  total  flux  per  pole  was  20.35  megalines.  What  is  the  true 
leakage  permeance  if  20  kiloampere-turns  were  required  at  that  flux  for 
the  air-gaps  and  the  armature,  per  pair  of  poles  ?  Ans.  234  perms. 

Prob.  12.  The  machine  specified  in  the  two  foregoing  problems 
requires  at  full  load  20  per  cent  more  ampere-turns  for  the  air-gap  and 
armature,  on  account  of  the  induced  voltage  being  12  per  cent  higher 
than  at  no-load.  The  armature  reaction  amounts  to  4000  demagnetizing 
ampere-turns  per  pole.  What  is  the  leakage  factor  at  full  load,  according 
to  the  calculated  leakage  permeance  and  according  to  that  obtained 
from  the  test?  Ans.  1.16;  1.19. 

Prob.  13.  A  closed  electric  circuit  consisting  of  a  battery  and  of 
a  bare  conductor  is  immersed  in  a  slightly  conducting  liquid,  so  that  part 
of  the  current  flows  through  the  liquid.  Indicate  the  common  points 
and  the  difference  between  this  arrangement  and  a  magnetic  circuit 
with  leakage.  Using  the  electrical  analogy,  show  that  armature  reaction 
increases  the  leakage  factor;  also  explain  the  fact  that,  in  order  to  com- 
pensate for  the  action  of  M  demagnetizing  ampere-turns  on  the  armature, 
more  than  M  additional  ampere-turns  are  required  on  the  pole-pieces. 

Prob.  14.  In  some  books  the  permeance  between  two  pole-faces 
(Fig.  30)  is  calculated  by  assuming  the  lines  of  force  to  be  concentric 
semicircles  as  shown  by  the  dotted  lines.  Show  that  such  a  permeance 
is  smaller  than  that  according  to  formula  (57)  and  therefore  should  not 
be  used.  Hint:  Compare  the  lengths  of  two  corresponding  lines  of 
force. 


CHAP. 'VI] 


EXCITING  AMPERE-TURNS 


115 


Prob.  15.  Let  AB  and  CD  (Fig.  31)  represent  the  cross-sections 
of  two  opposite  pole-faces  of  an  electromagnet,  inclined  at  an  angle  26 
to  one  another.  Show  that  of  the  three  assumptions  with  regard  to  the 
shape  of  the  lines  of  force  in  the  air  between  the  poles  (a)  is  more  correct 
than  (6)  and  (6)  is  more  correct  than  (c) ;  in  other  words,  the  assumption 
(a)  gives  a  higher  permeance  than  (6)  or  (c).  Hint:  tan  0  >  0  >  sin  0. 

Prob.  16.  Show  that  the  permeance  according  to  Fig.  31a,  between 
one  of  the  faces  and  the  plane  MN  of  symmetry,  is  equal  to 

(ph/fyLn(9w/l+l), 

where  h  is  the  width  of  the  pole-faces  perpendicular  to  the  plane  of  the 
paper,  and  that  formula  (56)  and  (57)  are  special  cases  of  it. 

Prob.  17.  The  formula  given  in  the  preceding  problem  is  deduced 
under  the  assumption  that  the  same  m.m.f.  is  acting  on  all  the  lines  of 


FIG.  31. — The  magnetic  paths  between  the  poles  of  an  electromagnet  (three 

assumptions). 

force.  Let  now  Fig.  31a  represent  a  cross-section  of  two  opposing  pole 
cores  in  an  electric  machine,  the  m.m.f.  between  A  and  C  being  zero, 
and  uniformly  increasing  to  a  value  M  between  the  points  B  and  D. 
Show  that  the  equivalent  permeance  of  the  path,  referred  to  the  m.m.f. 
M  is  equal  to  (tdi/d}[l-(l/6w)Ln(dw/l  +  l}]. 

NOTE.  If  it  is  desired  to  use  regularly  the  foregoing  formula  in  esti- 
mating the  leakage  factor,  the  values  of  the  expression  in  the  brackets 
[  ]  can  be  plotted  as  a  curve  for  the  values  of  (l/Ow)  as  abscissae.  Similar 
formulae  can  be  deduced  and  curves  plotted  for  the  permeance  of  the 
flank  leakage  between  adjacent  poles.  The  paths  of  the  lines  of  force 
over  the  poles  can  be  assumed  to  be  concentric  quadrants  and  between 
the  poles  to  have  a  shape  similar  to  that  indicated  in  Fig.  3 la. 

41.  The  Permeance  and  Reluctance  of  Irregular  Paths.     In 

using  the  methods  described  above  for  the  calculation  of  the. 
ampere-turns  for  the  air-gap,  the  teeth,  and  the  cores,  and  in  esti- 


116  THE  MAGNETIC  CIRCUIT  [ART.  41 

mating  the  leakage  factor,  the  reader  has  seen  the  difficulties 
involved  in  the  computation  of  the  permeance  of  an  irregular  path. 

In  the  parts  of  a  magnetic  field  not  occupied  by  the  exciting 
windings,  the  general  principle  applies  that  the  lines  of  force  and 
the  equipotential  surfaces  assume  such  shapes  and  directions  that 
the  total  permeance'  becomes  a  maximum,  or  the  reluctance  a 
minimum.  When  this  condition  is  fulfilled,  the  energy  of  the 
magnetic  field  becomes  a  maximum,  as  is  explained  in  Art.  57. 

When  the  field  needs  to  be  considered  in  two  dimensions  only, 
that  is,  in  the  case  where  we  have  long  cylindrical  surfaces  the 
properties  of  conjugate  functions  can  be  used  for  determining  the 
equations  of  the  lines  of  force  and  of  the  equipotential  surfaces; 
see  the  references  in  Art.  37  above.  However,  the  purely  mathe- 
matical difficulties  of  the  method  are  such  as  to  make  the  analytical 
calculation  of  permeances  feasible  in  the  simplest  cases  only. 

In  most  practical  cases,  especially  in  three-dimensional  prob- 
lems, recourse  must  be  had  to  the  graphical  method  of  trial  and 
approximation,  in  order  to  obtain  the  maximum  permeance. 
The  field  is  mapped  out  into  small  cells  by  means  of  lines  of  force 
and  equipotential  surfaces,  drawing  them  to  the  best  of  one's 
judgment;  the  total  permeance  is  calculated  by  properly  com- 
bining the  permeances  of  the  cells  in  series  and  in  parallel.  Then 
the  assumed  directions  are  somewhat  modified,  and  the  permeance 
is  calculated  again,  etc.,  until  by  successive  trials  the  positions  of 
the  lines  of  force  are  found  with  which  the  permeance  becomes  a 
maximum. 

The  work  of  trials  is  made  more  systematic  by  following  a  pro- 
cedure suggested  by  Lord  Rayleigh.  Imagine  infinitely  thin  sheets 
of  a  material  of  infinite  permeability  to  be  interposed  at  intervals 
into  the  field  under  consideration,  in  positions  approximately 
coinciding  with  the  equipotential  surfaces.  If  these  sheets  exactly 
coincided  with  some  actual  equipotential  surfaces,  the  total 
permeance  of  the  paths  would  not  be  changed,  there  being  no 
tendency  for  the  flux  to  pass  along  the  equipotential  surfaces.  In 
any  other  position  of  the  infinitely  conducting  sheets,  the  total 
permeance  of  the  field  is  increased,  because  through  these  sheets 
the  flux  densities  become  more  uniformly  distributed.  Moreover, 
these  sheets  become  new  equipotential  surfaces  of  the  system, 
because  no  m.m.f .  is  required  to  establish  a  flux  along  a  path  of 
infinite  permeance.  Thus,  by  drawing  in  the  given  field  a  system 


CHAP.  VI]  EXCITING  AMPERE-TURNS  117 

of  surfaces  approximately  in  the  directions  of  the  true  equipo- 
tential  surfaces,  and  assuming  these  arbitrary  surfaces  to  be  the 
true  equipotential  surfaces,  the  true  reluctance  of  the  path  is 
reduced.  In  other  words,  by  calculating  the  reluctances  of  the 
laminae  between  the  "  incorrect  "  equipotential  surfaces  and  adding 
these  reluctances  in  series,  one  obtains  a  reluctance  which  is  lower 
than  the  true  reluctance  of  the  path.  This  gives  a  lower  limit  for 
the  required  reluctance  (or  an  upper  limit  for  the  permeance)  of 
the  path. 

Imagine  now  the  various  tubes  of  force  of  the  original  field 
wrapped  up  in  infinitely  thin  sheets  of  a  material  of  zero  permeabil- 
ity. This  does  not  change  the  reluctance  of  the  paths,  because 
there  are  no  paths  between  the  tubes.  But  if  these  wrappings  are 
not  exactly  in  the  direction  of  the  lines  of  force,  the  reluctance  of 
the  field  is  increased,  because  the  densities  become  less  uniform, 
the  non-permeable  wrappings  forcing  the  lines  of  force  from  their 
natural  positions.  Thus,  by  drawing  in  a  given  field  a  system  of 
surfaces  approximately  in  the  directions  of  the  lines  of  force,  cal- 
culating the  reluctances  of  the  individual  tubes,  and  adding  them 
in  parallel,  a  reluctance  is  obtained  which  is  higher  than  the  true 
reluctance  of  the  path.  This  gives  an  upper  limit  for  the  reluc- 
tance (or  a  lower  limit  for  the  permeance)  of  the  path  under 
consideration. 

Therefore,  the  practical  procedure  is  as  follows:  Divide  the 
field  to  the  best  of  your  judgment  into  cells,  by  equipotential 
surfaces  and  by  tubes  of  force,  and  calculate  the  reluctance  of  the 
field  in  two  ways:  first,  by  adding  the  cells  in  parallel  and  the 
resultant  laminae  in  series;  secondly,  by  adding  the  cells  in  series 
and  the  resultant  tubes  in  parallel.  The  first  result  is  lower  than 
the  second.  Readjust  the  position  of  the  lines  of  force  and  of  the 
equipotential  surfaces  until  the  two  results  are  sufficiently  close  to 
one  another;  an  average  of  the  two  last  results  gives  the  true 
reluctance  of  the  field. 

One  difficulty  in  actually  following  out  the  foregoing  method 
is  that  the  changes  in  the  assumed  directions  of  the  field  that  will 
give  the  best  result  are  not  always  obvious.  Dr.  Th.  Lehmann  has 
introduced  an  improvement  which  greatly  facilitates  the  laying  out 
of  a  field.1  We  shall  explain  this  method  in  application  to  a  two- 

1  "  Graphische  Methode  zur  Bestiminung  des  Kraftlinienverlaufes  in  der 
Luft";  Ekktrotechnische  Zeitschrift,  Vol.  30  (1909),  p.  995. 


118  THE  MAGNETIC   CIRCUIT  [ART.  41 

dimensional  field,  though  theoretically  it  is  applicable  to  three- 
dimensional  problems  also.  According  to  Lehmann,  lines  of  force 
and  level  surfaces  are  drawn  at  such  distances  that  they  enclose 
cells  of  equal  reluctance.  Consider  a  slice,  or  a  cell,  in  a  two- 
dimensional  field,  v  centimeters  thick  in  the  third  dimension 
(where  v  =!//*),  and  of  such  a  form  that  the  average  length  Z  of 
the  cell  in  the  direction  of  the  lines  of  force  is  equal  to  its  average 
width  w  in  the  perpendicular  direction.  The  reluctance  of  such  a 
cell  is  always  equal  to  one  rel,  no  matter  whether  the  cell  itself  is 
large  or  small.  This  follows  from  the  fundamental  formula  for 
the  reluctance,  which  in  this  case  becomes  (R  =  vl/(vXw)  =  1. 

The  judgment  of  the  eye  helps  to  arrange  cells  of  a  width  equal 
to  the  length,  in  the  proper  position  with  respect  to  each  other  and 
to  the  adjoining  iron;  the  next  approximation  is  apparent  from 
the  diagram,  by  inspecting  the  lack  of  equality  in  the  average 
width  and  length  of  the  cells.  Lord  Rayleigh's  condition  is 
secured  by  this  means,  since  the  combination  of  cells  of  equal 
reluctance  leads  to  but  one  result,  whether  they  are  combined 
first  in  parallel  or  first  in  series.  After  a  few  trials  the  space  is 
properly  ruled,  and  it  simply  remains  to  count  the  number  of  cells 
in  series  and  in  parallel.  Dr.  Lehmann  shows  a  few  applications 
of  his  method  to  practical  cases  of  electrical  machinery,  and  the 
reader  is  referred  to  the  original  article  for  further  details. 

The  foregoing  methods  apply  only  to  the  regions  outside  the 
exciting  current,  because  only  in  such  parts  of  the  field  the  maxi- 
mum permeance  corresponds  to  the  maximum  stored  electromag- 
netic energy.  Within  the  space  occupied  by  the  exciting  windings 
the  condition  for  the  maximum  of  energy  is  different  (see  Art.  57), 
and  is  of  a  form  which  hardly  permits  of  the  convenient  application 
of  a  graphical  method.  However,  in  most  practical  cases  the 
directions  of  the  lines  of  force  within  the  exciting  windings  are 
approximately  known  a  priori:  or  else,  the  windings  themselves 
can  be  assumed,  for  the  purposes  of  computation,  to  be  concentrated 
within  a  very  small  space.  For  instance,  the  field  winding  can  be 
assumed  to  consist  of  an  infinitely  thin  layer  close  to  the  pole-waist. 
Then  the  condition  that  the  permeance  is  a  maximum  is  fulfilled  in 
practically  the  whole  field,  and  the  field  is  mapped  out  on  this  basis. 

Prob.  18.  Sketch  the  field  between  the  armature  and  a  pole-piece 
or  some  proportion  of  tooth,  slot,  and  air-gap  and  determine  the  lower 
and  upper  limits  of  the  reluctance  by  Lord  Rayleigh's  method. 


CHAP.  VI] 


EXCITING  AMPERE-TURNS 


119 


Prob.  19.  For  some  ratio  of  slot  width  to  air-gap  draw  the  tooth- 
fringe  field  to  the  perpendicular  surface  of  the  pole,  adjust  the  number 
and  spacing  of  the  lines  of  force  by  Dr.  Lehmann's  method,  and  see  how 
closely  you  can  check  the  corresponding  point  on  Carter's  curve  (Fig.  26). 

Prob.  20.  From  the  given  drawing  of  a  machine,  determine  the 
permeance  of  the  fringe  from  the  pole-tip  to  the  armature  by  Lehmann's 
method;  consult,  if  necessary,  Dr.  Lehmann's  original  article. 

Prob.  21.  Map  out  the  leakage  field  between  the  opposing  pole-tips 
and  cores  of  a  given  machine,  and  determine  its  equivalent  permeance  by 
Lehmann's  method,  assuming  the  field  coils  to  be  thin  and  close  to  the  core. 

41a.  The  Law  of  Flux  Refraction.  When  mapping  out  a  field 
in  air,  the  lines  of  force  must  be  drawn  so  as  to  enter  the  adjoining 
iron  almost  normally  to  its  surface,  even  if  they  are  continued  in 
the  iron  almost  parallel  to  its  surface.  The  lines  of  force  change 
their  direction  at  the  dividing  surface  suddenly  (Fig.  32),  and  in 
so  doing  they  obey  the  so-called  laio  of  flux  refraction;  namely, 


tan  di 


(58) 


Equipotential   \ 
surfaces        *\ 


Iron 


Air 


Since  /za  is  many  times  smaller  than  /*;,  the  angle  6a  is  usually  very 
small,  unless  Oi  is  very  nearly  90  degrees.  It  may  be  said  in  gen- 
eral that  the  lower  the  permeability 
of  a  medium  the  nearer  the  lines 
of  force  are  to  the  normal  at  its 
limiting  surfaces.  In  .this  way,  the 
path  between  two  given  points  is 
shortened  in  the  medium  of  lower 
permeability  and  is  lengthened  in 
the  medium  of  higher  permeability. 
Thus,  the  total  permeance  of  the 
circuit  is  made  a  maximum. 

To  deduce'  the  above-stated  law 
of  refraction,  consider  a  tube  of 
flux  between  the  equipotential  sur- 
faces ab  and  cd,  the  width  of  the 
path  in  the  direction  perpendicular 
to  the  plane  of  the  paper  being  one 
centimeter.  Let  Bi  and  Ba  be  the 
flux  densities,  and  Hi  and  Ha  the 

corresponding    magnetic    intensities    in   the   two    media.      Two 
conditions  must  be  satisfied,  namely,  first,  the  drop  of  m.m.f. 


FIG.  32.— The  refraction  of 
a  flux. 


120  THE  MAGNETIC  CIRCUIT  [ART.  41 

along  ac  is  the  same  as  that  along  bd,  and  secondly,  the  total 
flux  through  cd  is  equal  to  that  through  ab.     Or 


and 


Dividing  one  equation  by  the  other,  and  rearranging  the  terms, 
eq.  (58)  is  obtained. 

Prob.  22.  Show  that  the  total  refraction  which  is  in  some  cases 
experienced  by  rays  of  light  is  impossible  in  the  case  of  magnetic  lines 
of  force. 

Prob.  23.  Part  of  a  flux  emerges  from  the  flank  of  a  tooth  into  the 
slot  at  an  angle  of  1°  to  the  normal.  What  is  the  angle  which  the  lines 
of  force  make  with  the  side  of  the  slot  in  the  iron,  assuming  the  relative 
permeability  of  the  iron  to  be  1000?  Ans.  90°  -  0;  =  3°  IT. 


CHAPTER  VII 

THE  MAGNETOMOTIVE  FORCE  OF  DISTRIBUTED 

WINDINGS 

42.  The  M.M.F.  of  a  Direct-current  or  Single-phase  Dis- 
tributed Winding.  In  the  two  preceding  chapters  it  is  shown  how 
to  calculate  the  ampere-turns  required  for  a  given  flux  in  an  elec- 
tric machine.  When  the  exciting  winding  is  concentrated,  that  is, 
when  all  the  turns  per  pole  embrace  the  whole  flux,  the  number  of 
ampere-turns  is  equal  to  the  product  of  the  actual  amperes  flowing 
through  the  winding  times  the  number  of  turns.  Such  is  the  case 
hi  a  transformer,  in  a  direct-current  machine,  and  in  a  synchronous 
machine  with  salient  poles.  In  some  cases,  however,  the  exciting 
windings  are  distributed  along  the  air-gap,  so  that  only  a  part  of  the 
flux  is  linked  with  all  the  turns,  and  the  actual  ampere-turns  have 
to  be  multiplied  by  a  factor  in  order  to  obtain  the  effective  m.m.f . 
Such  is  the  case  in  an  induction  motor,  and  in  an  alternator  with 
non-salient  poles.  Moreover,  one  has  to  consider  the  m.m.f.  of 
distributed  armature  windings  when  calculating  the  performance 
of  a  machine  under  load,  because  the  armature  currents  modify  the 
no-load  flux.  In  this  chapter  the  m.m.fs.  of  distributed  windings 
are  treated  mainly  in  application  to  the  performance  of  the  induc- 
tion motor;  in  particular,  to  the  calculation  of  the  no-load  current 
and  the  reaction  of  the  secondary  currents.  The  armature  reaction 
in  synchronous  and  hi  direct-current  machines  is  analyzed  in  the 
next  two  chapters. 

Distributed  Winding  for  Alternator  Field.  A  cross-section  of  a 
four-pole  field  structure  with  non-salient  poles  for  a  turbo-alterna- 
tor is  shown  in  Fig.  33a.  The  flux  is  graded  (Fig.  336)  in  spite  of 
a  constant  air-gap,  because  the  total  ampere-turns  act  only  upon 
the  part  a  of  a  pole ;  two-thirds  of  the  ampere-turns  act  upon  the 
parts  6,  b  and  one-third  upon  the  parts  c,  c.  The  m.m.f.  and  the 
flux  in  the  parts  d,  d  are  equal  to  zero.  Thus,  theoretically,  the 
flux  density  in  the  air-gap  should  vary  according  to  a  "  stepped  " 

121 


122 


THE  MAGNETIC  CIRCUIT 


[ART.  42 


FIG.  33a.    A  four-pole  revolving  fieldstructure  with  non-salient  poles. 

curve  (Fig.  336) ;  in  reality,  the  corners  are  smoothed  out  by  the 
fringes.     The  total  number  of  ampere-turns  per  pole  must  be  such 

as  to  create  the  assumed 
maximum  flux  density 
Bmax  in  the  air-gap 
under  the  middle  part 
of  the  pole.  The  slots 
are  placed  with  due  re- 
gard to  the  mechanical 
strength  of  the  struc- 


-Pole-pitch- 


ture,  and  so  as  to  get  a 
flux-density      distribu- 
tion approaching  a  sine 
wave.     The  middle  part  of  the  curve  is  left  flat,  because  very  little 


FIG.  336.    The  flux-density  distribution  for 
the  field  shown  in  Fig.  33a. 


CHAP.  VII]         M.M.F.  OF  DISTRIBUTED  WINDINGS 


123 


flux  would  be  gained  by  placing  a  narrow  coil  near  the  center  of 
the  pole,  at  a  considerable  expense  in  copper,  and  in  power  loss  for 
excitation.  The  total  flux,  which  is  proportional  to  the  area  of  the 
curve  in  Fig.  336,  must  be  of  the  magnitude  required  by  eq.  (31) 
for  the  induced  e.m.f .  If  greater  accuracy  is  desired,  the  curve  in 
Fig.  336  is  resolved  into  its  fundamental  sine  wave  and  higher 
harmonics;  the  area  of  the  fundamental  curve  must  then  give 
the  flux  0  which  enters  into  eq.  (31). 

Single-phase  Distributed  Winding.  Let  us  consider  now  the 
stator  winding  of  an  induction  motor,  and  in  particular  the  m.m.f . 
created  by  the  current  in  one  phase.  We  begin  with  the  simplest 
case  of  a  winding  placed  in  one  slot  per  pole  per  phase  (Fig.  34). 
The  reluctances  of  the 
stator  core  and  of  the 
rotor  core  are  small  as 
compared  with  that  of 
the  air-gap  and  the 
teeth,  and  are  taken 
into  account  by  in- 
creasing the  reluctance 
of  the  active  layer  of 
the  machine  (air-gap 
and  teeth).  If  P and 
Q  are  the  centers  of 
the  slots  in  which  the 
opposite  sides  of  a  coil 
are  placed,  the  m.m.f. 
distribution  along  the  air-gap  is  that  shown  by  the  broken  line 
RPP'Q'QS.  In  other  words,  the  m.m.f.  across  the  active  layer, 
at  any  instant,  is  constant  over  a  pole  pitch,  and  is  alternately 
positive  and  negative  under  consecutive  poles. 

Let  n  be  the  number  of  turns  per  pole,  and  i  the  instanta- 
neous current ;  then  the  height  PP'  of  the  rectangle  is  equal  to  ni. 
It  is  understood  of  course  that  such  an  m.m.f.  acting  alone  does 
not  produce  the  sinusoidal  distribution  of  the  flux  density  assumed 
in  the  previous  chapters:  In  a  single-phase  motor  the  sinusoidal 
distribution  is  due  to  the  simultaneous  action  of  the  stator  and 
rotor  currents,  and  also  to  the  fact  that  the  windings  are  distrib- 
uted in  several  slots  per  pole.  In  a  polyphase  machine  the  simul- 
taneous action  of  the  two  or  three  phases  also  helps  to  secure  a 


FIG.  34. — The  m.m.f.  of  a  single-phase  unislot 
winding  resolved  into  its  harmonics. 


124  THE  MAGNETIC  CIRCUIT  [ART.  42 

sinusoidal  distribution.  As  long  as  the  coil  PQ  acts  alone,  the 
m.m.f.  has  a  "  rectangular  "  distribution  in  space,  and,  if  the  cur- 
rent in  the  coil  varies  with  the  time  according  to  the  sine  law,  the 
height  of  the  rectangle,  or  the  m.m.f.  across  the  active  layer,  also 
varies  according  to  the  sine  law.  In  what  follows  it  is  important 
to  distinguish  between  variations  of  the  m.m.fs.  in  space,  i.e.,  along 
the  air-gap,  and  those  occurring  in  time,  as  the  current  in  a  winding 
varies. 

For  the  purposes  of  analysis  the  rectangular  distribution  of  the 
m.m.f.  can  be  replaced  by  an  infinite  number  of  sinusoidal  distribu- 
tions (Fig.  34),  according  to  Fourier's  series.1  The  advantages  of 
such  a  development  over  the  orginal  rectangle  PP'Q'Q  are  as  fol- 
lows: 

(a)  The  sine  wave  is  a  familiar  standard  by  which  all  other 
shapes  of  periodic  curves  are  judged. 

(6)  When  adding  the  m.m.fs.  due  to  the  coils  in  different  slots, 
or  belonging  to  different  phases,  it  is  much  more  convenient  to  add 
sine  waves  than  to  add  rectangles  displaced  in  space  and  varying 
with  the  time. 

(c)  In  the  actual  operation  of  an  induction  motor  or  generator 
the  higher  harmonics  in  the  m.m.f.  wave  are  to  a  considerable 
extent  wiped  out  by  the  corresponding  currents  in  the  rotor,  so  that 
the  rectangular  distribution  is  actually  changed  to  a  nearly  sinu- 
soidal one  (see  Art.  45  below). 

Let  h  be  the  height  of  the  rectangle;  we  assume  that  for  all  the 
points  along  the  air-gap  the  sum  of  the  ordinates  of  all  the  sine 
waves  is  equal  to  h;  or 

Ti  =  Ai  sin  x+A3  sin  8^  +  ^.5  sin  5z+etc.  .     .     (59) 

Here  x  is  the  angle  in  electrical  degrees,  counted  along  the  air-gap, 
and  AI,  A 3,  A 5,  .  .  .  are  the  amplitudes  of  the  waves,  to  be  deter- 
mined as  functions  of  h.  No  cosine  harmonies  enter  into  this  for- 
mula, because  the  m.m.f.  distribution  is  symmetrical  with  respect 
to  the  center  line  00'  of  the  exciting  coil.  To  determine  the  ampli- 
tude of  the  nth  harmonic  An,  multiply  both  sides  of  eq.  (59)  by 
sin  nx  d(nx),  and  integrate  both  sides  between  the  limits  2=0  and 

1  For  the  general  method  of  expanding  a  periodic  function  into  a  series  of 
sines  and  cosines,  see  the  author's  Experimental  Electrical  Engineering,  Vol.  2, 
pp.  222  to  227. 


CHAP.  VII]         M.M.F.  OF  DISTRIBUTED  WINDINGS  125 

x  =  x.     All  the  terms  on  the  right-hand  side  vanish,  except  the  one 
containing  sin2  nx,  and  we  have 


from  which 

An=4h/(ra)  .......    .     (60) 

Thus,  the  required  series  is 

h  =  4/0r  (sin  x  +  J  sin  3x  +  £  sin  5z  +  etc.)     .     .     (61) 

This  means  that  the  amplitude  of  the  fundamental  wave  is  4/7: 
times  larger  than  the  height  h  of  the  original  rectangle;  the  ampli- 
tude of  the  third  harmonic  is  equal  to  one-third  of  that  of  the  fun- 
damental wave  ;  the  amplitude  of  the  fifth  harmonic  is  one-fifth  of 
that  of  the  fundamental  wave,  etc.  In  practical  applications  the 
fundamental  wave  is  usually  all  we  desire  to  follow,  but  in  some 
special  cases  a  few  of  the  harmonics  are  important.1 

Let  now  the  winding  of  a  phase  be  distributed  in  S  slots  per 
pole  (Figs.  15  and  16),  the  distance  between  the  adjacent  slots 
being  a  electrical  degrees.  The  conductors  in  every  pair  of  slots 
distant  by  a  pole  pitch  produce  a  rectangular  distribution,  of  the 
m.m.f  .  like  the  one  shown  in  Fig.  34,  or,  what  is  the  same,  an  equiv- 
alent series  of  sine-wave  distributions.  The  m.m.fs.  produced 
by  the  different  coils  are  superimposed,  and,  since  a  sum  of  sine 
waves  having  equal  bases  is  also  a  sine  wave,  the  resultant  m.m.f. 
also  consists  of  a  fundamental  sine  wave  and  of  higher  harmonics. 
The  fundamental  waves  of  the  m.m.fs.  of  the  several  coils  are  dis- 
placed by  an  angle  of  a  electrical  degrees  with  respect  to  one 
another,  so  that  the  amplitude  of  the  resultant  wave  is  not  quite 
S  times  larger  than  that  of  each  component  wave.  The  reduction 
coefficient,  or  the  slot  factor,  k8,  is  the  same  as  that  for  the  induced 
e.m.f.  (Art.  28),  because  in  both  cases  we  have  an  addition  of  sine 
waves  displaced  by  a  electrical  degrees,  (see  also  prob.  20  in  Art. 

1  This  method  of  treating  the  m.m.fs.  of  distributed  windings  by  resolv- 
ing the  rectangular  curve  into  its  higher  harmonics  is  due  to  A.  Blond  el. 
See  his  article  entitled  "  Quelques  proprietes  gene"rales  des  champs  magne"- 
tiques  tournants,"  L'Eclairage  Electrique,  Vol.  4  (1895),  p.  248.  Some  authors 
consider  the  actual  "  stepped  "  curves  of  the  m.m.f.  or  flux  distribution,  a 
procedure  rather  cumbersome,  and  in  the  end  less  accurate,  in  view  of  the 
fact  that  the  higher  harmonics  are  to  a  considerable  extent  wiped  out  by  the 
currents  in  the  rotor. 


126  THE  MAGNETIC  CIRCUIT  [ART.  42 

28.)  For  the  same  reason,  the  value  of  the  winding-pitch  factor, 
kw,  deduced  in  Art.  29,  holds  for  the  m.m.fs.  as  well  as  for  the 
induced  e.m.fs. 

When  adding  the  waves  of  the  higher  harmonics  due  to  several 
coils,  one  must  remember  that  an  angle  of  a  electrical  degrees  for 
the  fundamental  wave  is  equivalent  to  3a  electrical  degrees  for  the 
third  harmonic,  5c*  for  the  fifth  harmonic,  etc.  Therefore,  when 
using  the  formula  (29)  and  Fig.  19,  different  values  of  a  and  of  per 
cent  pitch  must  be  used  for  each  harmonic,  and  in  this  connection 
the  reader  is  advised  to  review  Art.  30.  In  the  practical  problems 
given  below  the  higher  harmonics  of  the  armature  m.m.f.  are  dis- 
regarded altogether.  The  results  so  obtained  are  in  a  sufficient 
agreement  with  the  results  of  experiments  to  warrant  the  great 
simplification  so  achieved.  For  the  completeness  of  the  treatment, 
and  as  an  application  of  the  general  method,  an  analysis  of  the 
effect  of  the  higher  harmonics  of  an  m.m.f.  is  given  in  Art.  45 
below.  However,  this  article  may  be  omitted,  if  desired,  without 
impairing  the  continuity  of  the  treatment  in  the  rest  of  the  book. 

Resolution  of  a  Pulsating  m.m.f.  into  Two  Gliding  m.m.fs.  The 
reader  is  aware  from  elementary  study  that  the  pulsating  m.m.fs. 
produced  by  two  or  three  phases  combine  into  one  gliding  (revolv- 
ing) m.m.f.  in  the  air-gap.  It  is  therefore  convenient  to  consider 
even  a  single-phase  pulsating  m.m.f.  as  a  combination  of  m.m.fs. 
gliding  along  the  air-gap  in  opposite  directions.  In  this  wise,  the 
m.m.fs.  due  to  different  phases  are  later  combined  in  a  simple 
manner.  This  method  of  treatment  is  similar  to  that  used  in 
mechanics,  when  an  oscillatory  motion  is  resolved  into  two  rotary 
motions  in  opposite  directions.  Also  in  the  analysis  of  polarized 
light  a  similar  method  of  treatment  is  used. 

Take  the  first  harmonic  of  the  m.m.f.  (Fig.  34)  and  assume  the 
current  in  the  exciting  coil  to  vary  with  the  time  according  to  the 
sine  law;  then  the  amplitude  of  the  m.m.f.  wave  also  varies  with 
the  time  according  to  the  sine  law.  Imagine  two  m.m.f.  waves,  of 
half  the  maximum  amplitude  of  the  pulsating  wave,  gliding  uni- 
formly along  the  air-gap  in  opposite  directions;  the  superposition 
of  these  waves  gives  the  original  pulsating  wave.  One  can  see 
this  by  drawing  such  waves  on  two  pieces  of  transparent  paper  and 
placing  them  in  various  positions  over  a  sketch  showing  the  pul- 
sating wave.  It  will  be  found  that  the  sum  of  the  corresponding 
ordinates  of  the  revolving  waves  gives  the  ordinate  of  the  pulsating 


CHAP.  VII]         M.M.F.  OF  DISTRIBUTED  WINDINGS  127 

wave  at  the  same  point.  Or  else,  represent  the  two  gliding  waves 
by  two  vectors  of  equal  magnitude  M,  revolving  in  opposite  direc- 
tions. The  resultant  vector  is  a  pulsating  one  in  a  constant  direc- 
tion, and  varies  harmonically  between  the  values  ±2M. 

The  analytical  proof  is  as  follows:  Let  the  exciting  current 
reach  its  maximum  at  the  moment  t= 0.  Then,  if  the  amplitude  of 
the  m.m.f.  wave  at  this  instant  is  equal  to  A,  the  amplitude  at  any 
other  instant  t  is  equal  to  A  cos  2nft.  Therefore,  the  m.m.f.  cor- 
responding to  a  point  distant  x  from  P  and  at  a  time  t  is  equal  to 
A  cos  2xft  sin  x.  By  a  familiar  trigonometrical  transformation 
we  have 

A  sin  z  cos  2rft=%A  sin  (x+2xft)  +%A  sin  (x  -2nft).       (62) 

The  right-hand  side  of  this  equation  represents  two  sine  waves,  of 
the  amplitude  %A,  gliding  synchronously  along  the  air-gap,  that  is, 
covering  one  pole  pitch  during  each  alternation  of  the  current. 
The  wave  %A  sin  (x+2nft)  glides  to  the  left,  because,  with  increas- 
ing t,  the  value  of  x  must  be  reduced  in  order  to  get  the  same  phase 
of  the  m.m.f.  wave,  that  is,  to  keep  the  value  of  (x+2nft)  constant. 
The  other  wave  glides  to  the  right,  because,  with  increasing  t,  the 
value  of  x  must  be  increased  in  order  to  obtain  any  constant  value 
of  (x—lnft).  A  similar  resolution  into  two  gliding  waves  can  be 
made  for  each  higher  harmonic  of  the  pulsating  m.m.f.  wave;  the 
higher  the  order  of  a  harmonic  the  lower  the  linear  speed  of  its  two 
gliding  wave  components. 

In  practice  it  is  usually  'required  to  know  the  relationship 
between  the  effective  value  i  of  the  magnetizing  current,  the  num- 
ber of  turns  n  per  pole  per  phase,  and  the  crest  value  of  one  of  the 
gliding  m.m.f.  waves.  From  the  preceding  explanation  this  rela- 
tionship for  the  fundamental  wave  is 

0.9A;fem,    ....     (63) 

where  M  is  the  amplitude  of  each  of  the  two  gliding  m.m.fs.,  niV2 
represents  the  maximum  height  h  of  the  original  rectangle,  and  the 
factor  J  is  introduced  because  the  amplitude  of  each  gliding  wave 
is  one-half  of  that  of  the  corresponding  pulsatingjwave.  The  breadth 
factor  kb  is  the  same  as  that  used  for  the  induced  e.m.fs.  (Arts.  27  to 
29) .  Similar  expressions  can  be  written  for  each  higher  harmonic, 
remembering  that  their  amplitudes  decrease  according  to  eq.  (61), 


128  THE  MAGNETIC  CIRCUIT  [ART.  43 

and  that  a  different  value  of  fa  must  be  used  for  each  harmonic. 
The  value  of  M  is  calculated  so  as  to  produce  the  required  revolv- 
ing flux,  as  is  explained  in  Chapters  IV,  V,  and  VI.  From  eq.  (63) 
either  n  or  i,  or  their  product  can  be  determined. 

Prob.  1.  A  single-phase  four-pole  induction  motor  has  24  stator  slots, 
two-thirds  of  which  are  occupied  by  the  winding;  there  are  18  con- 
ductors per  slot.  The  average  reluctance  of  the  active  layer  is  0.09  rel. 
per  square  centimeter.  What  current  is  necessary  to  produce  a  pulsating 
flux  of  such  a  value  that  the  maximum  flux  density  due  to  the  first 
harmonic  is  5  kl./sq.cm.,  when  the  secondary  circuit  is  open? 

Ans.    8.3  amp. 

Prob.  2.  Show  that  in  the  preceding  problem  the  difference  between 
the  actual  flux  per  pole  and  its  fundamental  is  less  than  2  per  cent. 

Prob.  3.  Show  that,  if  in  Fig.  34  the  angle  x  is  counted  from  the 
crest  of  the  first  harmonic,  the  expansion  into  the  Fourier  series  is  similar 
to  eq.  (61),  except  that  cosines  take  place  of  the  sines,  and  the  terms 
are  alternately  positive  and  negative. 

43.  The  M.M.F.  of  Polyphase  Windings.  Consider  a  two- 
phase  winding  of  the  stator  of  an  induction  motor  (Fig.  35a) ;  let 

f  2         Armature  i~ 

(66|  _  ftfj  [66]  JJ61 

p->-       °2^  slots  ^Ol      "*-; ; — • 

r\  i— 

FIG.  35a. — A  two-phase  winding. 

the  current  in  phase  1  lead  that  in  phase  2  by  \T,  or  by  90  electrical 
degrees.  A  little  reflection  will  show  that  the  resultant  m.m.f.  of 
the  two  phases  glides  from  right  to  left :  Let  the  current  in  phase  1 
reach  its  maximum  at  the  instant  £=0;  at  this  instant  the  current 
in  the  coil  2  is  zero,  and  the  m.m.f.  wave  is  distributed  uniformly 
under  the  coil  1 ;  at  the  instant  t= \T  the  current  in  phase  1  is  zero, 
and  the  m.m.f.  is  distributed  under  the  coil  2.  At  intermediate 
instants  both  coils  contribute  to  the  resultant  m.m.f.,  so  that  its 
maximum  occupies  a  position  intermediate  between  the  centers 
Oi  and  02  of  the  coils  1  and  2. 

The  actual  rectangular  distribution  of  the  m.m.f.  due  to  each 
phase  can  be  replaced  by  a  fundamental  sinusoidal  one  and  its 
higher  harmonics,  as  in  Fig.  34.  The  pulsating  fundamental  m.m.f. 
of  each  phase  can  be  replaced  by  two  waves  of  half  the  ampli- 
tude, gliding  synchronously  in  opposite  directions.  Let  the  wave 


CHAP.  VII]         M.M.F.  OF  DISTRIBUTED  WINDINGS  129 

due  to  phase  1,  and  gliding  to  the  left,  be  denoted  by  LI,  and  that 
due  to  phase  2  by  L2.  Let  the  corresponding  waves  gliding  to  the 
right  be  denoted  by  RI  and  R2.  Disregarding  the  higher  har- 
monics, the  resultant  m.m.f.  is  due  to  the  combined  action  of  the 
four  gliding  waves  L1?  L2,  RI  and  R2.  At  the  instant  t  =  Q  the 
crest  of  the  wave  LI  is  at  the  point  0\;  at  the  instant  t=\T  the 
crest  of  the  wave  L2  is  at  the  center  02  of  the  coil  2.  Consequently, 
at  the  instant  t=0  the  crest  of  the  wave  L2  is  90  electrical  degrees 
to  the  right  of  02,  or  it  is  at  0\.  Thus,  the  waves  LI  and  L2  actu- 
ally coincide  in  space,  and  form  one  wave  of  double  the  amplitude. 
The  crest  of  the  wave  RI  is  at  the  point  Oi  when  t=0;  the  crest 
of  R2  is  at  the  point  02  when  t=%T.  Therefore,  at  t=0  the  crest 
of  R2  is  90  electrical  degrees  to  the  left  of  the  point  02,  and  the 
waves  RI  and  R2  travel  at  a  distance  of  180  electrical  degrees  from 
each  other.  But  twro  such  waves  cancel  each  other  at  all  points 
and  at  all  moments,  so  that  there  is  no  resultant  R  wave.  Thus 
the  resultant  fundamental  wave  of  m.m.f.  in  a  two-phase  machine 
is  gliding.  Its  amplitude  is  twice  as  large  as  that  of  either  of  the 
component  gliding  m.m.fs.  of  the  two  phases,  which  components 

_  ,3  _  Armature        2  .\ 

|5S]  ifrjjj  J5S1  |S3|  |£fi~        J66]  [M]  [HT~ 


R 

FIG.  356.  —  A  three-phase  winding. 

are  expressed  by  eq.  (63).  If  the  current  in  phase  2  were  leading 
with  respect  to  that  in  phase  1,  the  L  fluxes  would  cancel  each 
other  and  the  resultant  flux  would  travel  from  left  to  right. 

Consider  now  a  three-phase  winding  (Fig.  356)  and  call  the 
m.m.fs.  which  glide  to  the  left,  and  which  are  due  to  the  separate 
phases,  by  LI,  L2,  and  L3  respectively.  Let  the  waves  which 
travel  to  the  right  be  denoted  by  RI,  R2,  and  R3.  Assume  the  cur- 
rent in  phase  2  to  be  lagging  by  120  electrical  degrees,  or  by  J77, 
with  respect  to  that  in  phase  1,  and  the  current  in  phase  3  to  be 
lagging  by  JT  with  respect  to  that  in  phase  2.  By  a  reasoning 
similar  to  that  given  for  the  two-phase  winding  above  it  can  be 
shown  that  the  three  L  waves  coincide  in  their  position  in  space, 
and  give  one  gliding  wave  of  three  times  the  amplitude  of  each 
wave.  The  three-  R  waves  are  relatively  displaced  by  240  elec- 


130  THE  MAGNETIC  CIRCUIT  [ABT.  43 

trical  degrees,  or,  what  is  the  same,  by  120  electrical  degrees; 
hence,  their  m.m.fs.  mutually  cancel  at  each  point  along  the 
air-gap.  This  can  be  proved  by  .drawing  three  sine  waves  dis- 
placed by  120  degrees  and  adding  their  ordinates  point  by  point; 
or  else  one  can  replace  each  wave  by  a  vector,  and  show  that  the 
sum  of  the  three  vectors  is  zero  because  they  form  an  equilateral 
triangle. 

The  reasoning  given  for  the  two-  and  three-phase  windings  can 
be  extended  to  any  number  of  symmetrical  phases,  say  m,  pro- 
vided that  the  windings  are  displaced  in  space  by  360/m  electrical 
degrees,  and  also  provided  that  the  currents  in  these  windings  are 
displaced  in  time  by  1/mth  of  a  cycle.  The  gliding  fundamental 
waves  due  to  each  phase  which  go  in  one  direction  are  in  phase 
with  each  other,  and,  when  added,  give  a  wave  m  times  larger 
than  that  expressed  by  eq.  (63)  ;  while  the  fundamental  waves 
going  in  the  opposite  direction  are  displaced  in  space  by  720/m 
electrical  degrees,  and  their  combined  m.m.f.  is  zero.  The  direc- 
tion in  which  the  resultant  m.m.f.  travels  is  from  the  leading  to 
the  lagging  phases  of  the  winding.  Thus,  for  any  symmetrical 
m-phase  winding 

.......     (64) 


w'here  M  denotes  the  amplitude  of  the  fundamental  sine  wave  of 
the  resultant  gliding  m.m.f.,  n  is  the  number  of  turns  per  pole  per 
phase,  and  i  is  the  effective  value  of  the  current  in  each  phase. 

Prob.  4.  It  is  desired  to  build  a  60  horse-power,  550-volt,  4-pole, 
Y-connected  induction  motor,  using  a  stator  punching  with  4  slots  per 
pole  per  phase,  and  a  winding  pitch  of  one  hundred  per  cent.  The  required 
maximum  m.m.f.  per  pole  is  estimated  at  1550  ampere-turns.  What  is 
the  total  required  number  of  stator  turns  (for  all  the  phases)  if  the  mag- 
netizing current  must  not  exceed  25  per  cent  of  the  full-load  current? 
The  estimated  full-load  efficiency  is  92  per  cent,  the  power  factor  at 
full  load  is  about  90  per  cent?  Ans.  Not  less  than  504. 

Prob.  5.  What  is  the  required  number  of  turns  in  the  preceding 
problem,  if  the  stator  winding  is  to  be  delta-connected  and  to  have  a 
winding  pitch  of  about  75  per  cent?  Ans.  Not  less  than  936. 

Prob.  6.  What  is  the  amplitude  of  the  first  harmonic  of  the  total 
armaturere  action  in  a  100-kva,  440-  volt,  6-pole,  two-phase  alternator 
with  non-salient  poles?  The  stator  has  72  slots;  the  coils  lie  in  slots  1 
and  9  ;  the  number  of  conductors  per  slot  is  Cs>  In  practice,  the  arma- 
ture reaction  must  not  exceed  a  certain  limit,  and  this  helps  to  determine 
the  permissible  value  of  Cs  Ans.  4800(7S  amp.  -turns. 


CHAP.  VII]         M.M.F.  OF  DISTRIBUTED  WINDINGS  131 

Prob.  7.  Plot  the  actual  "  stepped  "  curves  of  m.m.f.  distribution 
for  a  two-phase  winding  with  three  slots  per  pole  per  phase,  for  the  fol- 
lowing instants:  2  =  0,  t=^$T,  and  t  =  %T.  Compare  the  maximum  and 
the  average  m.m.f.  of  the  actual  distribution  with  those  of  the  first 
harmonic.1 

Prob.  8.  Solve  the  preceding  problem  for  a  three-phase  winding 
with  2  slots  per  pole  per  phase,  and  a  winding  pitch  of  £.  Take  two 
instants,  t=0  and  t=^T,  and  show  that  for  the  instants  t=^sTf 
etc.,  the  m.m.f.  distribution  is  the  same  as  for  t=Q,  while  for 
etc.,  the  m.m.f.  distribution  is  the  same  as  for  t  =  ^T. 

Prob.  9.  Prove  directly  that  two  equal  pulsating  sine  waves  of  m.m.f. 
or  flux,  displaced  by  90  electrical  degrees  in  space  and  in  time  relatively 
to  each  other,  give  a  gliding  sine  wave,  the  amplitude  of  which  is  equal 
to  that  of  each  pulsating  wave.  Solution:  The  left-hand  side  of  eq. 
(62)  gives  the  value  of  the  m.m.f.  at  a  point  x  and  at  an  instant  t,  due 
to  phase  1 ;  the  m.m.f.  produced  at  the  same  point  and  at  the  same  instant 
by  the  phase  2  is  A  sin  (x  +  %K)  cos  (2xft— %K).  Adding  the  two  expressions 
gives  A  sin  (x+2xft),  which  is  a  left-going  wave  of  amplitude  A. 

Prob.  10.  Prove,  as  in  the  preceding  problem,  that  the  three  pulsating, 
sine  waves  of  m.m.f.  produced  by  a  three-phase  winding,  give  together 
a  gliding  m.m.f.,  the  amplitude  of  which  is  50  per  cent  larger  than  that 
of  each  pulsating  wave. 

Prob.  11.  Prove  by  the  method  given  in  problem  9  above  that  m 
pulsating  m.m.f.  waves  displaced  in  space  and  in  tune  by  an  electrical 
angle  In/m  produce  a  gliding  m.m.f.  the  amplitude  of  which  is  %m  times 
larger  than  that  of  each  pulsating  wave.  See  Arnold,  Wechselstrom- 
technik,  Vol.  3  (1908)  p.  302. 

44.  The  M.M.Fs.  in  a  Loaded  Induction  Machine.2  Eq.  (64) 
gives  the  magnetizing  current  i  of  an  induction  motor  at  no-load, 
i.e.,  when  the  rotor  is  running  at  practically  synchronous  speed,  so 
that  the  secondary  currents  are  negligible.  When  the  motor  is 
loaded,  the  useful  flux  which  crosses  the  air-gap  is  due  to  the  com- 
bined action  of  the  primary  and  the  secondary  currents.  In  com- 
mercial motors  the  flux  at  full  load  is  but  a  few  per  cent  below  that 
at  no  load,  the  difference  being  due  to  the  impedance  drop  in  the 

1  Problems  7  and  8  are  intended  to  acquaint  the  student  with  the  usual 
method  of  calculation  of  the  m.m.fs.  of  distributed  windings  and  to  show  the 
advantage  of  Blondel's  method  used  in  the  text.     For  numerous  stepped 
curves  and  calculations,  see  Boy  de  la  Tour,  The  Induction  Motor,  Chapter  IV. 

2  The  treatment  in  this  article  presupposes  a  general  knowledge  of  the 
equivalent   performance   diagram   of   induction   machines;  the   purpose   of 
the  article  being  to  deduce  the  exact  numerical  relations.     This  article  and 
the  one  following  can  be  omitted  without  impairing  the  continuity  of  treat- 
ment in  the  rest  of  the  text. 


132  THE   MAGNETIC  CIRCUIT  [ART.  44 

primary  winding,  the  same  as  in  a  transformer.  Therefore,  the 
net  number  of  exciting  ampere-turns,  M,  is  approximately  the 
same  as  at  no  load.  This  means  that  the  geometric  sum  of 
the  m.m.fs.  produced  by  the  primary  and  the  secondary  currents  at 
any  load  is  nearly  equal  to  the  m.m.f .  due  to  the  primary  winding 
alone  at  no  load.  In  this  respect  the  induction  motor  is  similar  to 
a  transformer. 

(a)  Calculation  of  the  Secondary  Current.  Knowing  the- primary 
full-load  current,  the  secondary  full-load  current  can  be  calcu- 
lated from  the  required  counter-m.m.f . ;  the  procedure  can  be  best 
illustrated  by  an  example.  In  the  motor  given  in  prob.  4  above, 
the  full-load  current  is  estimated  at  57  amp. ;  taking  the  direction 
of  the  vector  of  the  applied  voltage  as  the  axis  of  reference,  the 
full-load  current  can  be  represented  as  51.3  —  /24.8  amp.  The 
magnetizing  current,  0.25X57=  14.25,  is  practically  in  quadrature 
with  the  applied  voltage,  because  it  is  hi  quadrature  with  the 
induced  counter  e.m.f .,  the  same  as  in  a  -transformer.  The  full- 
load  current  of  57  amp.  contains  a  component  which  supplies  the 
iron  loss  in  the  stator;  we  estimate  it  to  be  equal  to  about  1.1  a'mp. 
(2  per  cent  of  the  input).  Thus,  the  component  of  the  primary 
current,  the  action  of  which  must  be  compensated  by  the  second- 
ary currents,  is  '(51.3  -/24.8)  -  (1.1  -/14)  =  50.2  -j'10.8  amp.,  or  its 
absolute  value  is  5.14  amp.  This  is  called  the  current  transmitted 
into  the  secondary,  or  the  secondary  current  reduced  to  the  primary 
circuit.  This  current  produces  a  maximum  m.m.f.  of  0.9X3  XO. 958 
X42X51.4  =  5580  amp.-turns. 

Let  the  rotor  be  provided  with  a  three-phase  winding,  with  5 
slots  per  pole  per  phase,  and  let  the  winding  pitch  be  13/15.  The 
number  of  slots  is  selected  so  as  to  be  different  from  that  in  the  sta- 
tor, in  order  to  insure  a  more  uniform  torque,  and  to  reduce  the 
fluctuations  in  the  reluctance  of  the  active  layer.  We  have, 
according  to  eq.  (64),  that  5580  =  0.9X3X0.935  X  (ni),  from  which 
m  =  2210  amp.-turns.  Certain  practical  considerations,  for 
instance,  the  value  of  the  induced  secondary  voltage,  usually 
limit  the  choice  of  one  of  these  factors;  then  the  other  factor 
also  becomes  definite.  If,  for  instance,  the  rotor  is  to  have  10 
conductors  per  slot,  the  secondary  current  will  be  about  89  amp. 
The  secondary  i2r  loss  is  determined  by  the  desired  per  cent  slip; 
knowing  the  secondary  current  and  the  number  of  turns,  the 
necessary  size  of  the  conductor  can  easily  be  calculated. 


CHAP.  VII]        M.M.F.  OF  DISTRIBUTED  WINDINGS  133 

Sometimes  the  secondary  winding  consists  of  coils  individually 
short  circuited;  this  is  an  intermediate  type  of  winding  between 
an  ordinary  squirrel-cage  winding  and  a  three-phase  winding 
such  as  is  used  with  slip-rings.  Let  the  foregoing  motor  be 
provided  with  such  a  winding,  of  the  two-layer  type,  and  let  the 
rotor  have  71  slots,  6  conductors  per  slot,  the  coils  being  placed 
in  slots  1  and  14.  In  formula  (64)  m  stands  for  the  number  of 
symmetrically  distributed  phases,  the  current  in  each  phase  being 
displaced  in  time  by  2x/m  with  respect  to  that  in  the  next  phase. 
In  the  winding  under  consideration,  each  coil  represents  a  phase, 
and  one  has  to  go  over  a  pair  of  poles  until  one  finds  the  next  coil 
with  the  current  in  the  same  phase.  Thus,  in  this  case,  the  num- 
ber of  secondary  phases  is  equal  to  the  number  of  slots  per  pair  of 
poles,  or  m=35.5.  Each  coil  has  3  turns,  but  there  is  only  one  coil 
per  pair  of  poles,  so  that  n=1.5.  Substituting  these  values  into 
eq.  (64),  and  also  M  =  5580,  kb=  0.912,  we  find  i=  128  amp.  As  a 
matter  of  fact,  in  this  case  it  is  not  necessary  to  decide  what  the 
values  of  n  and  m  are,  because  eq.  (64)  contains  only  the  product 
mn,  which  is  the  total  number  of  turns  per  pole.  Thus,  in  our  case 
mn=(7lX3)/4. 

Formula  (64)  holds  also  for  a  squirrel-cage  winding,  the  number 
of  secondary  phases  being  equal  to  the  number  of  bars  per  pair  of 
poles.  Since  there  is  but  one  bar  per  phase,  each  bar  can  be  con- 
sidered as  one-half  of  a  turn,  and  in  formula  (64)  n=0.5  and  fc&=  1, 
so  that  it  becomes 


(64o) 


where  €2  is  the  total  number  of  rotor  bars,  and  p  is  the  number  of 
poles.  Or  else,  one  may  say  that  the  total  number  of  turns  per 
pole  is  equal  to  one-half  the  number  of  bars  per  pole,  so  that 
mn  =  %C2/p.  This  again  gives  eq.  (64a).  For  a  direct  proof  of 
formula  (64a)  see  problem  15  below.  Applying  this  formula  to 
the  same  rotor  with  71  slots  we  find  that  the  current  per  bar  is 
700  amp. 

(b)  The  Equivalent  Secondary  Winding  Reduced  to  the  Primary 
Circuit.  When  investigating  the  general  theory  of  the  induction 
motor  or  calculating  the  characteristics  of  a  given  motor,  it  is  con- 
venient to  replace  the  actual  rotor  winding  by  an  equivalent  wind- 
ing identical  with  the  primary  winding  of  the  motor.  In  this  case 
the  primary  current  transmitted  into  the  secondary  is  equal  to  the 


134  THE   MAGNETIC  CIRCUIT  [ART.  44 

actual  secondary  current  (one  to  one  ratio  of  transformation),  and 
the  primary  and  the  secondary  voltages  induced  by  the  useful  flux 
are  also  equal.  Each  electric  circuit  of  the  stator  then  can  be 
combined  with  the  corresponding  rotor  circuit.  In  this  manner 
the  so-called  "  equivalent  diagram  "  of  the  induction  motor  is 
obtained,1  a  way  of  representation  which  greatly  simplifies  the 
theory  of  the  machine. 

Let  12  be  the  secondary  current  in  the  coils  or  bars  of  the  actual 
rotor,  and  ir2  that  in  the  equivalent  rotor.  The  counter-m.m.f. 
of  both  rotors  must  be  the  same,  this  being  the  condition  of  their 
equivalence,  so  that 


from  which 

)  .....     (65) 


This  is  the  ratio  of  current  transformation  in  an  induction  motor. 
The  ratio  of  transformation  of  the  voltages  is  different,  namely, 


(66) 


In  an  ordinary  transformer  e2'/e2  =  12/^2  ==ni/n2,  because  there 
kbi  =  kb2=l,  and  m2  =  mi  =  l.  For  this  reason,  the  induction 
motor  is  sometimes  regarded  as  a  generalized  transformer. 

Taking  the  product  mie  for  the  actual  and  the  equivalent  rotor 
it  will  be  found  that  the  total  electric  power  input  is  the  same  in 
both,  provided  that  the  same  phase  displacement  is  preserved  in 
the  equivalent  rotor  as  in  the  original  one.  The  latter  condition  is 
essential  in  order  that  the  operating  characteristics  of  the  two 
machines  be  the  same.  This  means  (a)  that  the  total  i2r  loss  of 
the  equivalent  rotor  must  be  equal  to  that  of  the  original  rotor,  in 
order  to  preserve  the  same  slip,  and  (b)  that  the  leakage  react- 
ances of  the  two  rotors  must  affect  the  power  factor  of  the 
primary  current  in  the  same  way. 

Let  r2  and  r2  be  the  resistances  of  the  actual  and  of  the  equiva- 
lent rotor,  per  pole  per  phase.  We  have  the  condition  that 


(67) 


'Chas.  P.  Steinmetz,   Alternating    Current    Phenomena   (1908),   p.   249; 
Elements  of  Electrical  Engineering  (1905),  p.  263. 


CHAP.  VII]         M.M.F.   OF  DISTRIBUTED  WINDINGS  135 

Substituting  the  ratio  of  i2/i2  from  eq.  (65)  we  find 

r2/r2=(mi/m2)(kbinl/kb2n2).2      ....     (68) 
For  a  transformer  this  equation  reduces  to  the  familiar  expression 


The  ratio  of  the  inductances  is  the  same  as  that  of  the  resist- 
ances ;  this  can  be  proved  as  follows  :  In  order  that  the  equivalent 
winding  may  have  the  same  effect  on  the  power  factor  of  the  motor 
as  the  actual  winding,  the  equivalent  winding  must  draw  from  the 
line  an  equal  amount  of  reactive  volt-amperes,  due  to  its  leakage 
inductance.  The  magnetic  energies  stored  in  the  two  rotor  wind- 
ings must  therefore  be  equal,  and  we  have,  according  to  eq.  (104) 
in  Art.  58, 

ml.tf2'2L'2,       ......     (69) 


where  L2  and  Z/2  are  the  leakage  inductances  of  the  real  and  the 
equivalent  rotor  windings,  per  pole  per  phase.  The  form  of  this 
equation  is  the  same  as  that  of  eq.  (67)  ;  therefore,  substituting 
again  the  ratio  of  i'2/i2  from  eq.  (65)  an  expression  is  obtained  for 
the  ratio  of  L'2/L2  identical  with  that  given  by  eq.  (68),  namely 

L27L2=(m1/m2)(/c61n1//c62n2)2  .....     (70) 

This  result  could  also  be  foreseen  from  the  fact  that  the  reactances 
and  the  resistances  enter  symmetrically  in  the  equivalent  diagram, 
and  relation  (68)  holds  therefore  for  the  reactances  x2  and  x2. 
But  in  the  equivalent  diagram  the  secondary  and  the  primary  fre- 
quency is  the  same,  so  that  the  ratio  of  the  inductances  is  equal  to 
that  of  the  reactances;  this  gives  eq.  (70). 

It  must  be  clearly  understood  that  the  expressions  (68)  and  (70) 
refer  to  the  resistances  and  inductances  per  pole  per  phase.  When 
the  windings  of  a  phase  are  all  in  series,  both  in  the  stator  and  in  the 
rotor,  the  same  ratio  holds  of  course  for  the  resistances  per  phase  ; 
otherwise  the  actual  connections  must  be  taken  into  consideration, 
keeping  in  mind  that  the  total  i2r  loss  must  be  the  same  in  the 
equivalent  winding  as  in  the  actual  one.  Having  obtained  the  re- 
sistance of  the  equivalent  winding  per  pole,  the  turns  are  connected 
in  the  same  way  as  the  stator  turns.  This  fact  must  be  remembered 
in  particular  when  dealing  with  individually  short-circuited  coils 

1  See  the  author's  Experimental  Electrical  Engineering,  Vol.  2,  p.  77. 


136  THE  MAGNETIC  CIRCUIT  [ART.  45 

in  the  rotor,  or  with  a  squirrel-cage  winding.  In  these  two  cases 
the  individual  coils  or  bars  in  the  rotor  are  all  in  parallel,  while  the 
stator  coils  of  a  phase  are  usually  all  in  series,  or  in  two  parallel 
groups.  In  the  case  of  a  squirrel-cage  winding  the  resistance  r2 
includes  that  of  a  bar,  of  two  contacts  with  the  end-rings,  and  of 
the  equivalent  resistance  of  a  section  of  the  two  end-rings.1 

Prob.  12.  In  a  300  horse-power,  Y-connected,  14-pole  induction 
motor  the  full-load  current  is  estimated  to  be  310  amp.  The  primary 
winding  consists  of  336  turns  placed  in  168  slots;  the  winding-pitch  is 
0.75.  What  is  the  minimum  number  of  bars  in  the  squirrel-cage  second- 
ary winding,  if  the  current  per  bar  must  not  exceed  800  amp.?  The 
secondary  counter-m.m.f.  is  equal  to  about  90  per  cent  of  the  primary 
m.m.f.  Ans.  208. 

Prob.  13.  What  must  be  the  resistance  of  each  secondary  bar  in  the 
preceding  problem  (including  the  equivalent  resistance  of  the  adjoining 
segments  of  the  end-rings  and  also  of  the  contacts)  if  the  slip  at  full  load 
is  to  be  about  4  per  cent.?  Hint:  The  per  cent  slip  is  equal  to  the  i2r 
loss  in  the  rotor,  expressed  in  per  cent  of  the  power  input  into  the  second- 
ary. If  x  is  the  i2r  loss  in  the  rotor,  expressed  in  horse-power,  we  have 
that  x  =  0.04  (300  +  x) .  Ans.  70  microhms. 

Prob.  14.  The  motor  with  the  individually  short  circuited  second- 
ary coils,  that  is  used  as  an  illustration  in  the  text  above,  is  to  be 
investigated  with  respect  to  its  performance.  By  what  factor  must  the 
actual  resistance  and  inductance  of  each  secondary  coil  be  multiplied 
in  order  to  obtain  the  equivalent  resistance  and  inductance  per  primary 
phase?  Also  by  what  factor  must  the  equivalent  current  be  multiplied 
in  order  to  obtain  the  actual  current  in  each  secondary  coil? 

Ans.  297;  0.402. 

Prob.  15.  Prove  formula  (64o)  directly,  by  considering  the  m.m.f s. 
of  the  individual  bars.  Solution :  At  any  instant  the  currents  in  the  bars 
under  a  pole  are  distributed  in  space  according  to  the  sine  law,  because 
the  gliding  flux  which  jnduces  these  currents  is  sinusoidal.  The  average 
current  per  bar  is  *A/2X(2/*)  =0.9i.  The  number  of  turns  per  pole  is 
C2/2p,  and  all  these  turns  are  active  at  the  crest  of  the  m.m.f.  wave. 
Therefore,  Jlf  =  0.9i(C2/2p). 

45.  The  Higher  Harmonics  of  the  M.M.FS.  In  the  preceding 
study,  the  effect  of  the  higher  harmonics  in  the  m.m.f.  wave  was  dis- 
regarded. In  fact,  these  harmonics  usually  exert  a  negligible 
influence  upon  the  operation  of  a  good  polyphase  induction  motor, 
under  normal  conditions.  These  m.m.f.  harmonics  move  at  lower 
speeds  than  the  fundamental  field ;  therefore,  the  fluxes  which  they 

1  See  the  author's  Electric  Circuit;  also  E.  Arnold,  Die  Wechselstromtechnik, 
Vol.  5,  Part  I  (1909),  p.  57. 


CHAP.  VII]        M.M.F.  OF  DISTRIBUTED  WINDINGS  137 

produce  cut  the  secondary  conductors  at  comparatively  high  rela- 
tive speeds ;  thus,  secondary  currents  are  induced  which  wipe  out 
these  harmonics  to  a  considerable  degree.  There  are  practical  cases, 
however,  in  which  some  one  particular  harmonic  becomes  of  some 
importance,  and  affects  the  operation  of  the  machine,  particularly 
at  starting.  For  this  reason  the  following  general  outline  of  the 
properties  of  the  higher  harmonics  in  the  m.m.f .  is  given.1 

In  a  single-phase  machine  (Fig.  34)  all  the  higher  harmonics  of 
the  m.m.f.  are  pulsating  at  the  same  frequency  as  the  fundamental 
wave,  but  the  width  of  the  nth  harmonic  is  only  I/nth  of  that  of 
the  fundamental  wave.  Each  pulsating  harmonic  can  be  replaced 
by  two  gliding  harmonics  of  half  the  amplitude,  one  left-going,  the 
other  right-going.  The  linear  velocity  of  these  gliding  m.m.fs.  is 
only  I/nth  of  that  of  the  fundamental  gliding  waves,  because  they  ~ 
cover  in  the  time  %T  a  distance  equal  only  to  their  own  base,  PQ/n  \ 
(180  electrical  degrees) .  With  one  slot  per  pole,  the  amplitudes  of 
the  higher  harmonics  decrease  according  to  eq.  (61),  but  with  more 
than  one  slot,  or  with  a  fractional-pitch  winding  they  decrease 
more  rapidly,  because  different  values  of  kb  must  be  taken  for  each 
harmonic  (see  Art.  30  above). 

In  a  two-phase  machine,  consider  (Fig.  35a)  the  gliding  waves 
Ln  and  Rn,  of  the  nth  harmonic.  For  this  harmonic,  the  distance 
between  0\  and  0%  is  equal  to  \nn  electrical  degrees.  At  the 
instant  t—0  the  crest  of  the  wave  Lnl  is  at  the  point  0\\  at  the 
instant  t=\T  the  crest  of  the  wave  Ln2  is  at  the  point  0^.  There- 
fore, the  two  waves  travel  at  a  relative  distance  of  \n(n  —  1)  elec- 
trical degrees,  considering  the  base  of  the  nth  harmonic  as  equal 
to  its  own  180  electrical  degrees.  In  a  similar  manner,  the  distance 
between  the  crests  of  the  two  right-going  waves  is  found  to  be 
equal  to  %x(n  + 1)  electrical  degrees.  We  thus  obtain  the  following 
table  of  the  angular  distances  between  the  waves  due  to  the  two 
phases : 

Order  of  the  harmonic 1       3        5        7        9         11         13 

Distance  between  the  two  Ln  waves        0       it         In       %K       IK       5;r         GTT 
Distance  between  the  two  Rn  waves        K       2^       STT       4rc       5?r       6n         IK 

The  waves  which  travel  at  a  distance  0,  2n,  4;r,  etc.,  are  simply 
added  together,  while  those  at  a  distance  TT,  3nf  5n,  etc.,  cancel  each 

1  For  a  more  detailed  treatment  see  Arnold,  Wechselstromtechnik,  Vol. 
3  (1904),  Chapter  13,  and  Vol.  5,  part  I  (1909),  Chapter  9. 


138  THE  MAGNETIC   CIRCUIT  [ART.  45 

other.  Thus,  in  a  two-phase  machine,  the  3d,  7th,  llth,  etc., 
harmonics  travel  against  the  direction  of  the  main  m.m.f.,  while  the 
5th,  9th,  13th,  etc.,  harmonics  travel  in  the  same  direction  as  the 
fundamental  m.m.f.,  though  at  lower  peripheral  speeds. 

Applying  a  similar  reasoning  to  a  three-phase  winding  (Fig. 
356)  we  find  that  the  three  Ln  waves  travel  at  a  relative  distance  of 
|7r(n — 1),  while  the  relative  distance  between  the  three  Rn  waves  is 
f x(n+ 1)  electrical  degrees.  We  thus  obtain  the  following  table  of 
the  angular  distances  between  the  waves  due  to  the  three  phases : 

Order  of  the  harmonic 1      3      5      7      9     11    13     15 

Distance  between  the  three  Ln  waves. ..    0      IT:    |TT    0     tyc  *px    0   ^-n 
Distance  between  the  three  Rn  waves. . .     |TT    f  n     0    V^  *£*    0    ^-K  ^TT 

The  component  waves,  of  any  harmonic,  which  travel  at  a  distance 
zero  from  each  other,  are  simply  added  together,  and  give  a  resul- 
tant wave  of  three  times  the  amplitude  of  the  component.  The 
three  waves  which  travel  at  an  angular  distance  of  f  n  or  one  of 
its  multiples  from  each  other  give  a  sum  equal  to  zero.  Thus,  in  a 
three-phase  machine,  the  1st,  7th,  13th,  etc.,  harmonics  travel  in 
one  direction,  while  the  5th,  llth,  17th,  etc.,  harmonics  travel 
against  the  direction  of  the  fundamental  m.m.f.  The  higher  the 
,  order  of  a  harmonic  the  lower  its  peripheral  speed.  The  harmonics 
of  the  order  3,  9,  15,  etc.,  are  entirely  absent. 

Prob.  16.  What  are  the  amplitudes  of  the  fifth  and  the  seventh 
harmonics,  in  percentage  of  that  of  the  fundamental  wave,  for  a  three- 
phase  winding  placed  in  2  slots  per  pole  per  phase,  when  the  winding- 
pitch  is  5/6?  Ans.  1.4  and  1.0  per  cent  respectively. 

Prob.  17.  Show  that,  in  order  to  eliminate  the  nth  harmonic  in 
the  m.m.f.  wave,  the  winding-pitch  must  satisfy  this  condition ;  namely, 
f/i:  =  (2q  + 1)  /n,  where  f  is  defined  in  Fig.  16,  and  q  is  equal  to  either  0, 
1,  2,  3,  etc.  Hint:  Cos  \fn  must  be  =  0. 

Prob.  18.  Investigate  the  direction  of  motion  of  the  various  har- 
monics of  the  m.m.f.  in  a  symmetrical  w-phase  system. 

Prob.  19.  Show  that  only  the  nth  harmonic  in  the  m.m.f.  wave, 
due  to  the  nth  harmonic  in  the  exciting  current,  moves  synchronously 
with  the  fundamental  gliding  m.m.f.,  and  therefore  distorts  it  perma- 
nently. 

Prob.  20.  A  poorly  designed  2-phase,  60-cycle  induction  motor  has 
4  poles,  1  slot  per  phase  per  pole,  and  a  winding  pitch  of  100  per  cent. 
At  what  sub-synchronous  speed  is  it  most  likely  to  stick?  Hint:  The 
torque  due  to  any  harmonic  reverses  as  the  motor  passes  through  the 
corresponding  sub-synchronous  speed.  Anp.  360  r.p.m. 


CHAPTER  VIII 


ARMATURE   REACTION   IN   SYNCHRONOUS 
MACHINES 

46.  Armature  Reaction  and  Armature  Reactance  in  a  Syn- 
chronous Machine.  When  a  synchronous  machine  carries  a  load, 
either  as  a  generator  or  as  a  motor,  the  armature  currents,  being 
sources  of  m.m.f.,  modify  the  flux  created  by  the  field  coils,  and 
thus  influence  the  performance  of  the  machine.  Fig.  36  shows  an 


•<* Direction  of  Rotation         <m& 

FIG.  36 — The  flux  distribution  in  a  single-phase  synchronous  machine 
under  load. 

instantaneous  flux  distribution  in  the  simplest  case  of  a  single- 
phase  alternator,  with  one  slot  per  pole;  the  armature  conductors 
are  marked  a  and  b.  With  the  directions  of  the  armature  and  field 
currents  indicated  in  the  sketch,  the  flux  is  crowded  toward  the 
right-hand  tips  of  the  poles.  In  order  to  show  this,  imagine  two 
fictitious  conductors  of  and  V  with  currents  equal  and  opposite 

139 


140  THE  MAGNETIC  CIRCUIT  IART.  46 

to  those  in  the  actual  conductors  a  and  b  respectively.  The  addi- 
tion of  these  fictitious  conductors  does  not  modify  the  armature 
m.m.f.  because  they  neutralize  each  other.  The  conductor  a' 
may  be  considered  as  forming  a  turn  with  a,  while  bf  forms  a  turn 
with  b.  It  will  be  seen  that  the  m.m.f.  of  the  coil  aaf  assists  that 
of  the  field  coil  A,  while  the  m.m.f.  of  the  coil  W  is  opposite  to  that 
of  the  field  coil  B. 

The  armature  current  in  the  coil  ab  not  only  distorts  the  no- 
load  field,  but  also  reduces  the  total  flux  per  pole.  This  may  be 
seen  by  considering  the  flux  in  the  four  parts  of  the  air-gap,  marked 
x,  y,  x',  and  y',  where  x  =  x'  and  y  =  y'-  The  sum  of  the  fluxes  in 
the  portions  y  and  yr  is  the  same  as  without  the  armature  current, 
because  the  flux  density  in  the  part  y  is  increased  by  the  same 
amount  by  which  it  is  reduced  in  the  part  y'  (neglecting  satura- 
tion) .  But  in  the  parts  x  and  xf  the  flux  is  reduced  by  the  arma- 
ture m.m.f.,  so  that  the  total  result  over  the  pole-pitch  is.  a  reduc- 
tion in  the  value  of  the  no-load  flux.  The  position  of  the  armature 
conductors  and  the  direction  of  the  armature  currents  have  been 
selected  arbitrarily.  They  can  be  chosen  so  that  the  flux  will  be 
crowded  toward  the  left-hand  tips  of  the  poles,  or  so  that  the  total 
flux  will  be  increased  by  the  armature  m.m.f.,  instead  of  being 
reduced.  The  influence  of  the  armature  currents,  in  modifying  the 
value  of  the  field  flux  and  distorting  it,  is  called  the  armature  reac- 
tion. The  armature  reaction  is  measured  in  ampere-turns,  since 
it  is  a  magnetomotive  force. 

In  addition  to  the  general  distortion  of  the  field  by  the  arma- 
ture currents,  there  is  a  local  distortion  around  each  armature 
conductor.  This  distortion  does  not  extend  into  the  pole  shoes, 
but  is  limited  to  the  slots  and  the  air-gap;  it  is  indicated  in  Fig.  36 
by  ripples  in  the  flux  around  a  and  b.  These  ripples  may  be 
regarded  as  a  result  of  the  superposition  upon  the  main  flux  of  the 
local  fluxes  4>a  and  <P&  excited  by  the  armature  currents.  While 
these  local  fluxes,  shown  by  the  dotted  lines,  have  no  real  existence, 
except  around  the  end  connections  of  the  armature  conductors, 
it  is  convenient  to  consider  them  separately.  They  are  purely 
alternating  fluxes,  in  phase  with  the  currents  with  which  they  are 
linked,  so  that  they  induce  in  the  armature  windings  alternating 
e.m.fs.  in  a  lagging  phase  quadrature  with  the  currents. 

The  effect  of  these  local  fluxes  upon  the  voltage  of  the  machine 
is  represented  by  a  certain  armature  reactance,  because  the  effect  is 


CHAP.  VIII]    REACTION  IN  SYNCHRONOUS  MACHINES  141 

the  same  as  if  the  armature  winding  created  no  leakage  fluxes 
around  it,  but  a  separate  reactance  coil  were  connected  in  series 
with  each  armature  lead.  The  calculation  of  the  armature  react- 
ance, or  of  the  local  fluxes,  is  treated  in  Art.  67,  the  subject  of  this 
chapter  being  armature  reaction  only,  that  is,  the  effect  of  the  load 
upon  the  main  magnetic  circuit.  In  the  numerical  problems  of  this 
chapter,  for  the  solution  of  which  it  is  necessary  to  know  the  value 
of  the  armature  reactance,  this  value  is  given.  It  is  not  quite 
correct,  strictly  speaking,  to  separate  the  local  distortion  of  the 
main  flux  as  a  phenomenon  by  itself;  moreover,  the  separation  is 
somewhat  indefinite  and  arbitrary.  However,  the  flux  so  separated 
is  comparatively  small,  and  the  treatment  of  the  armature  reaction 
proper  is  thereby  greatly  simplified. 

The  distribution  shown  in  Fig.  36  varies  from  instant  to  instant 
because  the  relative  position  of  the  armature  changes  with  refer- 
ence to  the  poles,  as  well  as  the  value  of  the  armature  current. 
Besides,  there  are  usually  two  or  three  armature  phases,  and  sev- 
eral slots  per  pole  per  phase.  It  would  be  out  of  the  question  to 
calculate  the  actual  fluxes  for  each  instant  and  to  take  into  account 
their  true  influence  upon  the  e.m.f.  induced  in  the  armature.  In 
practice,  certain  approximate  average  values  of  armature  reaction 
and  of  armature  reactance  are  employed,  which  permit  one  to 
predict  the  actual  performance  of  a  machine  with  a  sufficient 
accuracy. 

In  the  case  of  a  synchronous  generator  (alternator)  the  problem 
usually  presents  itself  in  the  following  form :  It  is  required  to  pre- 
determine the  field  ampere-turns  necessary  for  a  prescribed  ter- 
minal voltage  at  a  given  load.  Knowing  the  resistance  and  the 
leakage  reactance  of  the  armature,  the  voltage  drop  in  the  arma- 
ture is  added  geometrically  to  the  terminal  voltage;  this  gives  the 
induced  voltage  in  the  machine.  Knowing  from  the  no-load  satura- 
tion curve  the  required  net  excitation  at  this  voltage,  and  correct- 
ing it  for  the  effect  of  the  armature  reaction,  the  necessary  field 
ampere-turns  are  obtained.  The  results  of  such  calculations  for 
different  values  of  the  armature  current  and  for  various  power 
factors,  plotted  as  curves,  are  called  the  load  characteristics  of  the 
alternator. 

In  the  case  of  a  synchronous  motor  the  terminal  voltage  is  usu- 
ally given,  and  it  is  required  to  determine  the  field  excitation  such 
that,  at  a  given  mechanical  output,  the  input  to  the  armature  be  at 


142  THE  MAGNETIC  CIRCUIT  [ART.  46 

a  given  power-factor;  a  leading  power-factor  is  usually  prescribed, 
in  order  to  raise  the  lagging  power-factor  of  the  whole  plant.  The 
problem  is  solved  in  like  manner  to  that  of  the  generator,  by 
taking  into  account  the  proper  signs  when  calculating  the  reactance 
drop  and  the  armature  reaction.  The  results,  plotted  in  the  form 
of  curves,  are  called  the  phase  characteristics,  or  V-curves  of  a 
synchronous  motor.1 

It  will  be  seen  from  Fig.  36  that  the  crowding  of  the  flux  to  one 
pole-tip,  by  the  armature  currents,  is  primarily  due  to  the  fact  that 
the  pgles  shown  there  are  projecting  or  salient,  so  that  the  reluc- 
tance along  the  air-gap  is  variable.  With  non-salient  poles  the 
flux  is  simply  shifted  sidewise  without  being  distorted.  Therefore, 
before  going  into  the  details  of  the  calculation  of  armature  reaction 
in  machines  with  salient  poles  we  shall  first  consider  (in  the  next 
article)  the  case  of  a  machine  with  non-salient  poles. 

^r 

Prob.  1.  Draw  the  distribution  of  the  flux,  similar  to  that  shown 
in  Fig.  36,  when  the  armature  conductors  are  opposite  the  centers  of  the 
poles,  and  when  they  are  somewhere  between  the  adjacent  pole-tips. 

Prob.  2.  Explain  the  details  of  the  flux  distribution  in  Fig.  36,  by 
means  of  a  hydraulic  analogy,  assuming  A  and  B  to  represent  two  main 
centrifugal  pumps,  and  a  and  b  to  be  two  smaller  pumps  placed  in  the 
stream. 

Prob.  3.  Let  each  field  coil  in  Fig.  36  have  N  turns,  and  let  the 
exciting  current  be  /;  let  the  number  of  conductors  at  a  be  Cs,  and  the 
instantaneous  value  of  the  armature  current  i.  What  is  the  total  flux 
per  pole,  if  the  average  permeance  of  the  machine  per  pole  is  (P  perms 
per  electrical  radian,  and  the  angles  6  and  x  are  in  electrical  radians? 

Ans.  (2NI6-Csix)(P.  in  maxwells. 

Prob.  4.  Let  a  synchronous  machine  be  loaded  in  such  a  way  that 
the  armature  current  reaches  its  maximum  when  the  conductors  a  and  b 
(Fig.  36)  are  opposite  the  centers  of  the  poles,  in  other  words,  the  current 
is  in  phase  with  the  e.m.f.  which  would  be  induced  at  no  load.  Prove 
that  (neglecting  saturation)  the  average  flux  per  pole  during  a  complete 
cycle  is  the  same  as  without  the  armature  reaction,  but  is  crowded  to 
the  leading  tip  of  the  pole,  i.e.,  in  the  direction  of  rotation  in  the  case  of 
a  motor,  and  to  the  trailing  tip,  or  against  the  direction  of  rotation  when 
the  machine  is  working  as  a  generator.  Hint:  The  flux  is  weakened  as 
much  in  the  position  x  of  the  conductors  as  it  is  strengthened  in  the 
symmetrical  position  x'',  the  distortion  is  in  the  same  direction  in  both 
positions. 

!See  the  author's  "  Experimental  Electrical  Engineering,"  Vol.  2,  p.  121; 
also  his  "  Essays  on  Synchronous  Machinery,"  General  Electric  Review, 
1911,  p.  214. 


CHAP.  VIII]    REACTION  IN  SYNCHRONOUS  MACHINES  143 

Prob.  6.  Let  a  synchronous  machine  be  loaded  in  such  a  way  that 
the  armature  current  reaches  its  maximum  when  the  conductors  a  and  6 
(Fig.  36)  are  midway  between  the  poles,  in  other  words,  when  the  current 
is  displaced  by  90  electrical  degrees  with  respect  to  the  e.m.f.  induced 
at  no  load.  Prove  that  the  average  distortion  during  a  complete  cycle 
is  zero,  but  that  the  flux  is  weakened  if  the  armature  current  lags  behind 
the  induced  e.m.f.,  and  is  strengthened  by  a  leading  current.  Hint: 
The  flux  is  weakened  in  both  of  the  symmetrical  positions,  x  and  xr,  of 
the  conductors,  but  the  distortion  is  in  opposite  directions. 

Prob.  6.  In  a  single-phase  synchronous  machine  the  armature 
current  reaches  its  maximum  when  the  armature  conductors  are  dis- 
placed by  an  angle  ^  with  respect  to  the  centers  of  the  poles;1  prove  that 
the  field  is  distorted  by  the  component  i  cos  </>  of  the  current  and  is 
weakened  or  strengthened  by*  the  component  i  sin  ^. 

47.  The  Performance  Diagram  of  a  Synchronous  Machine  with 
Non-Salient  Poles.  Let,  in  a  machine  with  non-salient  poles,  the 
field  winding  be  placed  in  several  slots  per  pole,  so  that  the  field 
m.m.f .  in  the  active  layer  of  the  machine  is  approximately  distrib- 
uted according  to  the  sine  law.  Consider  the  machine  to  be  a 
polyphase  generator  supplying  a  partly  inductive  load.  The  ampli- 
tude of  the  first  harmonic  of  the  armature  reaction  has  the  value 
given  by  eq.  (64)  in  Art.  43,  and  revolves  synchronously  with  the 
field  m.m.f.,  as  is  explained  there.  Since  the  sum  of  two  sine  waves 
is  also  a  sine  wave,  the  resultant  m.m.f.  is  also  distributed  in  the 
active  layer  of  the  machine  according  to  the  sine  law. 

To  deduce  the  phase  displacement,  in  space,  between  the  two 
sine  waves,  consider  the  coil  a  b  (Fig.  36)  to  be  one  of  the  phases  of 
the  polyphase  armature  winding.  For  reasons  of  symmetry,  the 
maximum  m.m.f.  produced  by  a  polyphase  winding  is  at  the  center 
of  the  coil  in  which  at  that  particular  moment  the  current  is  at  a 
maximum.  Assume  first  that  the  current  in  the  phase  a  b  reaches  its 
maximum  when  the  conductors  a  and  b  are  opposite  the  centers  of 
the  poles.  The  maximum  armature  m.m.f.  at  that  instant  is  dis- 
placed by  90  electrical  degrees  with  respect  to  the  center  lines 
of  the  poles.  The  direction  of  the  armature  current  is  determined 
by  the  well-known  rule,  and  it  is  found  to  be  such  that  the  arma- 
ture m.m.f.  lags  behind  that  of  the  pole,  considering  the  direction 
of  rotation  of  the  poles  as  positive.  Since  both  m.m.fs.  revolve 
synchronously,  this  angle  between  the  two  m.m.f.  crests  is  pre- 

1  The  angle  $  is  different  from  the  external  phase-angle  0  between  the 
current  and  the  terminal  voltage;  see  Fig.  37. 


144  THE  MAGNETIC  CIRCUIT  [ART.  47 

served  all  the  time.  Thus,  in  a  polyphase  generator,  the  armature 
m.m.f.  lags  behind  the  field  m.m.f.  by  90  electrical  degrees  in 
space,  when  the  currents  are  in  phase  with  the  voltages  induced  at 
no-load.  This  statement  is  in  accord  with  that  in  problem  4  in  the 
preceding  article,  because,  if  each  phase  shifts  the  flux  against  the 
direction  of  rotation,  all  the  phases  together  simply  increase  the 
result. 

Let  now  the  currents  in  the  armature  windings  be  lagging 
90  electrical  degrees  behind  the  corresponding  e.m.fs.  induced 
at  no-load.  This  simply  means  that  the  armature  m.m.f.  is  shifted 
further  back  by  90  degrees  as  compared  to  the  case  considered 
before;  therefore,  the  angle  between 'the  field  m.m.f.  and  the 
armature  m.m.f.  is  180  electrical  degrees,  and  the  two  m.m.fs.  are 
simply  in  phase  opposition.  This  is  in  accord  with  the  statement 
in  prob.  5. 

From  the  two  preceding  cases  it  follows  that,  when  in  a  syn- 
chronous machine  with  non-salient  poles  the  currents  lag  by  an 
angle  </>  electrical  degrees  (Figs.  37  and  38)  with  respect  to  the 
induced  voltage  at  no-load,  the  armature  m.m.f.  wave  lags  by  an 
angle  of  90  +  ^  electrical  degrees  behind  the  field  m.m.f.  wave.  In 
the  case  of  a  generator  with  leading  currents  the  angle  (p  is  negative ; 
in  a  synchronous  motor  <p  is  larger  than  90  degrees. 

Let,  in  Fig.  37,  i  be  the  vector  of  the  current  in  one  of  the  phases, 
and  let  e  be  the  corresponding  terminal  voltage,  the  phase  angle 
between  the  two  being  <£.  Adding  to  e  in  the  usual  way  the  ohmic 
drop  ir  in  the  armature,  in  phase  with  i,  and  the  reactive  drop  ix 
in  leading  quadrature  with  i,  the  induced  voltage  E  in  the  same 
phase  is  obtained.1  The  resultant  useful  flux,  $,  which  induces 
this  e.m.f.  leads  E  by  90  degrees  in  time;  0  is  in  phase  with  the 
net  or  resultant  m.m.f.  Mn  which  produces  it.  The  m.m.f.  Mn  is  a 
sum  of  the  field  m.m.f.  Mf  and  of  the  armature  reaction  Ma 

1  On  account  of  skin  effect  and  eddy  currents  in  the  armature  conductors, 
the  effective  resistance  r  to  alternating  currents  is  considerably  higher  than  that 
calculated  or  measured  with  direct  current.  The  actual  amount  of  increase 
depends  upon  the  character  of  the  winding,  the  size  of  the  conductors,  the 
shape  of  the  slots,  the  frequency,  etc.,  so  that  no  definite  rule  can  be  given. 
Fortunately,  the  ohmic  drop  constitutes  but  a  small  percentage  of  the  voltage 
of  a  machine,  so  that  a  considerable  error  committed  in  estimating  the  value 
of  the  ir  drop  affects  the  voltage  relations  but  very  little.  See  A.  B.  Field, 
"Eddy  Currents  in  Large  Slot-wound  Conductors,"  Trans.  Amer.  Inst. 
Elect.  Engrs.,  Vol.  24  (1905),  p.  761. 


CHAP.  VIII]    REACTION  IN  SYNCHRONOUS  MACHINES 


145 


expressed  by  eq.  (64) .  The  triangle  OFG  represents  the  relations  in 
space,  while  the  figure  OABD  is  a  time  diagram.  Therefore,  the 
two  figures  are  independent  of  one  another;  but  it  is  convenient 
to  combine  them  into  one,  by  using  the  common  vectors  0  and  i. 


FIG.  37. — The  performance  diagram  of  a  synchronous  generator,  with 
non-salient  poles. 

With  respect  to  the  triangle  OFG,  the  vector  i  represents  the  posi- 
tion of  the  crest  of  the  armature  m.m.f .  relatively  to  the  crest  OG 
of  the  field  m.m.f.,  the  angle  between  the  two  being  90  +  </>,  as  is 
explained  above.  Thus,  the  vector  Ma  is  in  phase  with  i. 


146  THE  MAGNETIC  CIRCUIT  [ART.  47 

When  i  and  e  are  given,  the  vector  E  is  easily  found  if  the 
resistance  and  the  reactance  of  the  armature  winding  are  known. 
The  required  net  excitation,  Mn,  is  then  taken  from  the  no-load 
saturation  curve  of  the  machine,  and  Ma  is  figured  out  from  eq. 
(64).  Then  the  required  field  ampere-turns,  Af/,  are  found  from 
the  diagram,  either  graphically  or  analytically. 

The  diagram  shown  in  Fig.  37  is  known  as  the  Potier  diagram. 
Strictly  speaking,  it  is  correct  only  for  machines  with  non-salient 
poles,  but  as  an  approximate  semi-empirical  method  it  is  some- 
times used  for  machines  with  projecting  poles,  in  place  of  the  more 
correct  diagram  shown  in  Fig.  40.  Fig.  37  represents  the  condi- 
tions in  the  case  of  a  generator  with  lagging  currents.  When  the 
current  is  leading  the  vector  i  is  drawn  to  the  left  of  the  vector  e, 
with  the  corresponding  changes  in  the  other  vectors. 

A  similar  diagram  for  a  synchronous  motor  which  draws  a 
leading  current  from  the  line  is  shown  in  Fig.  38.  The  vector  e' 
represents  the  line  voltage,  and  e  is  the  equal  and  opposite  voltage 
which  is  the  terminal  voltage  of  the  machine  considered  as  a  gen- 
erator. The  rest  of  the  diagram  is  the  same  as  in  Fig.  37.  A  lead- 
ing current  with  respect  to  the  line  voltage  e'  is  a  lagging  current 
with  respect  to  the  generator  terminal  voltage  e,  so  that  the  field 
is  weakened  by  the  armature  reaction  in  both  cases  (Mn  <  Mf  in 
both  figures).  The  energy  component  i\  of  the  current  is  reversed 
in  the  motor,  therefore  the  field  is  shifted  in  the  opposite  direc- 
tion; Mn  leads  Mf  in  the  motor  diagram  and  lags  behind  it  in  the 
generator  diagram.  The  case  of  a  synchronous  motor  with  a 
lagging  current  can  be  easily  analyzed  by  analogy  with  the  above- 
described  cases. 

In  practice,  it  is  usually  preferred  to  represent  the  relations 
shown  in  Figs.  37  and  38  analytically,  rather  than  to  actually  con- 
struct a  diagram.  The  following  relations  hold  for  both  the  gen- 
erator and  the  motor.  Projecting  all  the  sides  of  the  polygon 
OABD  on  the  direction  e  and  on  the  direction  perpendicular  to 
and  leading  e  by  90  degrees,  we  have 

E  cos(f)g=e+ir  cos<f)+ix  sin  <£,       .     .     .     (71) 
E  sin  <j)K=ix  cos<£—  ir  sin  <£,       ....     (72) 

where  cf>e  is  the  angle  between  the  vectors  e  and  E,  counted  positive 
when  E  leads  e,  as  in  Fig.  37.     The  subscript  z  suggests  that  the 


CHAP.  VIII]     REACTION  IN  SYNCHRONOUS  MACHINES 


147 


angle  </>2  is  due  to  the  impedance  of  the  armature.  The  expressions 
i  cos  $  and  i  sin  0  represent  the  energy  component  and  the  react- 
ive component  of  the  current  respectively;  they  are  designated  in 


FIG.  38. — The  performance  diagram  of  a  synchronous  motor,  with 
non-salient  poles. 

Figs.  37  and  38  by  ii  and  i2.     Denoting  the  right-hand  sides  of  the 
eqs.  (71)  and  (72)  by  ei  and  e2  for  the  sake  of  brevity,  we  have: 


62  =  i\x  —i2r. 


(73) 
(74) 


148  THE  MAGNETIC  CIRCUIT  [ART.  47 

Squaring  eqs.  (71)  and  (72)  and  adding  them  together  gives 

'~  (75) 


Dividing  eq.  (72)  by  (71)  results  in 

tan</>2=e2Ai  .......     (76) 

Consequently,  the  angle  between  E  and  i  becomes  known;  namely, 

.......     (76a) 


where  </>'  is  called  the  internal  phase  angle.  Knowing  E,  the  cor- 
responding excitation  Mn  is  taken  from  the  no-load  saturation 
curve  of  the  machine  ;  from  the  triangle  OFG  we  have  then  : 


asm<f>',       .     .     .     (77) 

where  </>'  is  known  from  eq.  (76a)  .  In  numerical  applications  it  is 
convenient  to  express  all  the  M's  in  kiloampere-turns. 

The  diagram  shown  in  Fig.  38  and  the  equations  developed 
above  can  be  used  for  determining  not  only  the  phase  characteris- 
tics of  a  synchronous  motor,  but  its  overload  capacity  at  a  given 
field  current  as  well.  This  latter  problem  is  of  extreme  importance 
in  the  design  of  synchronous  motors.  The  input  into  the  machine, 
per  phase,  is  —  ei  cos  <£>  ;  the  part  ir  of  the  line  voltage  is  lost  in  the 
armature,  the  part  ix  corresponds  to  the  magnetic  energy  which  is 
periodically  stored  in  the  machine  and  returned  to  the  line,  without 
performing  any  work.  The  remainder,  E,  corresponds  to  the  use- 
ful work  done  by  the  machine,  plus  the  iron  loss  and  friction.  If 
the  armature  possessed  no  resistance  and  no  leakage  reactance  the 
terminal  voltage  would  be  equal  to  E  in  magnitude  and  in  phase 
position.  Thus,  the  expression  —  Ei  cos  <f>',  corrected  for  the  core 
loss  in  the  armature  iron,  represents  the  input  into  the  revolving 
structure,  per  phase.  The  overload  capacity  of  the  machine  is 
determined  by  the  possible  maximum  of  this  expression. 

The  problem  is  complicated  by  the  fact  that  the  relation 
between  E  and  Mn  is  expressed  by  the  no-load  saturation  curve, 
which  is  difficult  to  represent  by  an  equation.  The  problem  is 

1  In  numerical  applications  it  is  more  convenient  to  use  the  approximate 
formula 


l   ..........     (75a) 

obtained  by  the  binomial  expansion  of  expression  (75)  ;  since  all  other  terms 
can  be  neglected  when  e2  is  small  as  compared  to  et. 


CHAP.  VIII]    REACTION  IN  SYNCHRONOUS  MACHINES  149 

therefore  solved  by  trials,  assuming  a  certain  reasonable  value  of 
E,  and  calculating  the  expression  —  Ei  cos  (/>',  until  a  value  of  E 
is  found,  for  which  this  expression,  corrected  for  the  core  loss,  fric- 
tion, and  windage,  becomes  a  maximum.  The  problem  of  finding 
i  and  (f>f  for  an  assumed  E  is  a  definite  one,  because  the  four  equa- 
tions (71),  (72),  (76a)  and  (77)  contain  only  four  unknown  quanti- 
ties, i,  <f>,  <j>zj  and  </>'.  Instead  of  solving  the  problem  by  trials,  an 
analytical  relation  can  be  assumed  between  E  and  Mn,  on  the  use- 
ful part  of  the  no-load  saturation  curve,  for  instance  a  straight  line 
(not  passing  through  the  origin),  a  parabola,  etc.  The  problem  is 
then  solved  by  equating  the  first  derivative  of  the  product — 
Ei  cos  ft  to  zero,  having  previously  expressed  E,  i,  and  cos  <j>' 
through  some  one  independent  variable.  Both  methods  have 
been  worked  out  for  a  synchronous  motor  with  salient  poles.1  The 
relations  are  simplified  for  a  machine  with  non-salient  poles. 

The  foregoing  theory  of  the  armature  reaction  does  not  apply 
directly  to  single-phase  machines.  The  pulsating  armature  reac- 
tion in  such  a  machine  can  be  resolved  into  two  revolving  reactions, 
as  in  Art.  42.  The  reaction  which  revolves  in  the  same  direction 
with  the  main  field  is  taken  into  account  as  in  a  polyphase  machine. 
The  inverse  reaction  is  partly  wiped  out  by  the  eddy  currents  pro- 
duced in  the  metal  parts  of  the  revolving  structure ;  it  is  therefore 
difficult  to  express  the  effect  of  this  reaction  theoretically.  The 
treatment  in  this  book  is  limited  to  polyphase  machines,  which  are 
used  in  practice  almost  exclusively.2 

Prob.  7.  In  the  100  kva.,  440-volt,  6-pole,  two-phase  alternator, 
given  in  Problem  6,  Art.  43,  the  amplitude  of  the  first  harmonic  of  the 
armature  reaction  was  4800(7S  ampere-turns.  What  is  the  per  cent 
voltage  regulation  of  the  machine  at  a  power-factor  of  80  per  cent 
lagging,  if  C«  =  l,  that  is  if  the  armature  has  one  conductor  per  slot? 
The  armature  reactance  is  0.038  ohm,  and  the  armature  resistance  is 
0.008  ohm,  both  per  phase.  The  no-load  saturation  curve  of  the 
machine  is  as  follows : 

e  =  400    440    490     525     550    volts. 
Mn=6.7     8.0    10.0    12.0    14.0    kiloamp. -turns. 

Ans.     22  per  cent. 

1  See  the  author's  "Essays  on  Synchronous  Machinery,"  General  Electric 
Review,  1911,  July  and  September. 

2  In  regard  to  the  armature  reaction  in  single-phase  machines,  see  E. 
Arnold,  Die    Wechselstromtechnik,  Vol.  4  (1904),  pp.~  32-39;    Pichelmayer, 
Dynamobau  (1908),  pp.  251-259;    Max  Wengner,  Theoretische  und  Expert- 
mentelle  Untersuchungen  an  der  Synchronen  Einphasen-Maschine  (Oldenbourg, 
1911.) 


150  THE  MAGNETIC  CIRCUIT  [ART.  48 

Prob.  8.  The  machine  specified  above  is  to  be  used  as  a  synchronous 
motor.  Determine  graphically  the  required  field  excitation  when  the 
useful  output  on  the  shaft  is  to  be  700  kw.,  and  in  addition  the  machine 
must  draw  from  the  line  600  leading  reactive  kva.  The  efficiency  of  the 
machine  at  the  above-mentioned  load  is  estimated  to  be  about  91  per 
cent.  Ans.  12.5  kiloampere-turns. 

Prob.  9.  Draw  to  the  same  scale  as  the  diagram  shown  in  Fig.  37, 
another  similar  diagram,  for  the  same  value  of  the  current  and  of  the 
phase  angle  0,  except  that  the  current  is  to  be  leading.  Assume  a 
reasonable  shape  of  the  saturation  curve  in  determining  the  new  value  of 
Mn.  Show  that  a  much  smaller  exciting  current  is  required  with  the 
same  kva.  output,  than  in  the  case  of  a  lagging  current. 

Prob.  10.  Solve  problem  9  for  the  motor  diagram  shown  in  Fig.  38, 
assuming  the  current  to  be  lagging  with  respect  to  the  line  voltage. 

Prob.  11.  For  a  given  alternator,  show  how  to  determine  the  voltage 
e  (Fig.  37),  analytically  or  graphically,  when  Mf,  i,  and  ^  are  given; 
explain  when  such  a  case  arises  in  practice. 

Prob.  12.  For  a  given  synchronous  motor,  show  how  to  determine 
the  reactive  component  i2  of  the  current  (Fig.  38),  analytically  or  graphic- 
ally, when  Mf,  e  and  it  are  given ;  explain  when  such  a  case  arises  in 
practice. 

Prob.  13.  Work  out  the  details  of  the  above-mentioned  method 
for  the  determination  of  the  overload  capacity  of  a  synchronous-  motor 
by  trials.  Hint:  Introduce  the  components  of  fc  and  i,  in  phase  and  in 
quadrature  with  E;  rewrite  eqs.  (71)  and  (72)  by  projecting  the  figure 
OABD  on  the  direction  of  E  and  on  that  perpendicular  to  E.  Use  no 
angles  in  the  formulae,  and  neglect  the  small  terms  containing  r,  where 
they  lead  to  complicated  equations  of  higher  degrees. 

48.  The  Direct  and  Transverse  Armature  Reaction  in  a  Synchro- 
nous Machine  with  Salient  Poles.  In  a  machine  with  non-salient 
poles  the  armature  reaction  shifts  the  field  flux  but  hardly  distorts 
its  shape.  In  a  machine  with  projecting  poles  the  flux,  generally 
speaking,  is  both  altered  in  value  and  crowded  toward  one  pole- 
tip  (Fig.  36).  It  is  convenient,  therefore,  to  resolve  the  traveling 
wave  of  the  armature  m.m.f .  into  two  waves,  one  whose  crests  coin- 
cide with  the  center  lines  of  the  poles,  the  other  displaced  by  90 
electrical  degrees  with  respect  to  it.  The  first  component  of  the 
armature  m.m.f.  produces  only  a  "  direct  "  effect  upon  the  field 
flux,  that  is,  it  either  strengthens  or  weakens  the  flux,  without  dis- 
torting it.  The  second  component  produces  a  "  transverse " 
action  only,  viz.,  it  shifts  the  flux  toward  one  or  the  other  pole-tip, 
without  altering  its  value  (that  is,  neglecting  the  saturation) . 

We  have  seen  before  that  an  armature  current,  which  reaches 
its  maximum  when  the  conductor  is  opposite  the  center  of  the 


CHAP.  VIII]    REACTION  IN  SYNCHRONOUS  MACHINES  151 

pole,  distorts  the  flux;  while  a  current  in  quadrature  with  the 
former  exerts  a  direct  reaction  only.  It  is  natural,  therefore,  to 
resolve  the  actual  current  in  each  phase  into  two  components,  in 
time  quadrature  with  each  other,  and  in  such  a  way  that  each 
component  reaches  its  maximum  in  one  of  the  above-mentioned 
principal  positions  of  the  conductor  with  respect  to  the  field-poles. 
Let  the  current  in  each  phase  be  i,  and  let  it  reach  its  maximum  at 
an  angle  </>  after  the  induced  no-load  voltage  is  a  maximum  (Fig. 
40) .  Then,  the  two  components  of  the  current  are 

id=i  sin  (p 
and 

it  —  i  cos  <[>. 

The  component  id  produces  a  direct  armature  reaction  only, 
and  the  component  it  a  transverse  reaction  only.1 

For  practical  calculations,  and  in  order  to  get  a  concrete  picture 
of  the  armature  reaction,  it  is  convenient  to  represent  the  armature 
reaction  as  shown  in  Fig.  39.  Namely,  the  direct  reaction,  due  to 
the  components  id  of  the  armature  currents,  is  replaced  by  an  equiv- 
alent number  of  concentrated  ampere-turns  Md  on  the  pole.  The 
value  of  Md  is  selected  so  that  its  action  in  reducing  or  strength- 
ening the  flux  is  equal  to  the  true  action  of  the  armature  currents. 
The  transverse  reaction,  due  to  the  component  it  of  the  armature 
currents,  is  replaced  by  a  certain  number  of  ampere-turns,  Mt,  on 
the  fictitious  poles,  (£),  (N),  shown  by  dotted  lines  between  the  real 
poles.  For  simplicity,  and  for  other  reasons  given  in  Art.  51,  the 
fictitious  poles  are  assumed  to  be  of  a  shape  identical  with  that  of 
the  real  poles.  The  number  of  exciting  ampere-turns  Mt  is  so 
chosen,  that  the  effect  of  the  fictitious  poles  is  approximately  the 
same  as  that  of  the  distorting  ampere-turns  on  the  armature. 

The  flux  of  the  fictitious  poles  strengthens  the  flux  of  the  real 
poles  on  one  side  and  weakens  it  by  the  same  amount  on  the  other 
side,  so  that  the  fictitious  poles  actually  distort  the  main  flux 
without  altering  its  value.  Strictly  speaking,  the  complete  action 
of  the  distorting  ampere-turns  on  the  armature  cannot  be  imitated 

1  The  resolution  of  the  armature  reaction  in  a  synchronous  machine 
into  a  direct  and  a  transverse  reaction  was  first  done  by  A.  Blondel.  See 
V Industrie  Electrique,  1899,  p.  481 ;  also  his  book  Moteurs  Synchrones  (1900), 
and  two  papers  of  his  in  the  Trans.  Intern,  Electr.  Congress,  St.  Louis,  1904, 
Vol.  1,  pp.  620  and  635. 


152 


THE  MAGNETIC  CIRCUIT 


[ART.  48 


by  fictitious  poles  of  the  same  shape  as  the  main  poles,  because 
harmonics  of  appreciable  magnitude  are  thereby  neglected.  How- 
ever, actual  experience  shows  that  the  performance  of  a  machine, 
calculated  in  this  way,  can  be  made  to  check  very  well  with  the 
observed  performance,  by  properly  selecting  the  coefficients  of  the 
direct  and  the  transverse  reaction.  In  a  generator,  the  flux  is 
crowded  against  the  direction  of  rotation  of  the  poles  (Fig.  36) ; 
consequently,  the  fictitious  poles  lag  behind  the  real  poles,  as 
shown  in  Fig.  39.  In  a  synchronous  motor  they  lead  the  real 
poles  by  90  electrical  degrees. 

If  the  ratio  of  the  pole  arc  to  pole-pitch  were  equal  to  unity, 
as  with  non-salient  poles,  the  whole  wave  of  the  demagnetizing 


Actual  distribution 
of  transverse  flux 


Flux  distribution 
due  to  fictitious  pole 


FIG.  39.  —  The  direct  and  transverse  armature  reactions  in  a  synchronous 
machine,  represented  by  fictitious  poles  and  field  windings. 

m.m.f.  of  the  armature  would  be  acting  upon  the  pole,  and  the 
equivalent  concentrated  m.m.f.  Md  on  the  pole  would  have  to  be 
equal  to  the  average  value  of  the  actual  distributed  armature 
m.m.f.  We  would  have  then 


(78) 


where  the  maximum  armature  m.m.f.  is  determined  by  eq.  (64), 
Art.  43,  and  2/7r=0.637  is  the  ratio  of  the  average  to  the  maximum 
ordinate  of  a  sine  wave.  In  reality,  only  a  part  of  the  armature 
m.m.f.,  the  one  near  its  amplitude,  acts  upon  the  poles,  the  action 
of  lower  parts  of  the  wave  being  practically  zero  because  of  the 
gaps  between  the  poles.  Therefore,  the  ratio  between  the  maxi- 
mum m.m.f.  M  sin  <p  and  the  average  equivalent  m.m.f.  M^  is 


CHAP.  VIII]    REACTION   IN  SYNCHRONOUS  MACHINES  153 

larger  than  0.637.  For  the  ordinary  shapes  of  projecting  poles, 
experiment  and  calculation  (see  Art.  50  below)  show  that  this 
ratio  varies  between  0.81  and  0.85.  Using  an  average  of  these 
limits  instead  of  2/7r  in  eq.  (78)  and  substituting  for  M  its  expres- 
sion from  eq.  (64)  we  obtain  the  following  practical  formula  for 
estimating  the  armature  demagnetizing  ampere-turns  per  pole  in  a 
synchronous  machine  with  projecting  poles: 


(79) 


In  this  formula  i  sin  (j>  is  the  component  id  of  the  armature  current, 
per  phase.  In  actual  machines  the  numerical  coefficient  in  this 
formula  varies  between  0.73  and  0.77,  depending  on  the  shape  of 
the  poles  and  the  ratio  of  pole-arc  to  pole-pitch. 

By  a  similar  reasoning,  if  the  ratio  of  pole-arc  to  pole-pitch  were 
equal  to  unity,  the  equivalent  number  of  exciting  ampere-turns  on 
the  fictitious  poles  would  be 

3f«=(2/7r)Mcos#  ......     (80) 


Since  the  ratio  of  pole-arc  to  pole-pitch  on  the  fictitious  poles 
is  less  than  unity,  the  numerical  coefficient  should  be  larger  than 
2/7T.  But,  on  the  other  hand,  the  permeance  of  the  air-gap  under 
the  fictitious  poles  is  much  higher  than  the  actual  permeance  of 
the  machine  in  the  gaps  between  the  poles,  so  that  a  much  smaller 
number  of  ampere-turns  Mt  is  sufficient  to  produce  the  same 
distorting  flux.  The  combined  effect  of  these  two  factors  is  to 
reduce  the  coefficient  in  formula  (80)  to  a  value  considerably 
below  2/7T.  For  the  usual  shapes  of  projecting  poles,  experiment 
and  calculation  (See  Art.  51  below)  show  that  this  ratio  varies 
between  0.30  and  0.36.  Using  an  average  of  these  limits  instead 
of  2/7T  hi  eq.  (80),  and  substituting  for  M  its  expression  from  eq. 
(64),  we  obtain  the  following  practical  formula  for  estimating 
the  distorting  ampere-turns  per  pole,  in  a  synchronous  machine 
with  projecting  poles: 


(81) 


In  this  formula  i  cos  ^  is  the  component  it  of  the  armature  cur- 
rent, per  phase.  In  some  actually  built  machines  the  coefficient 
in  this  formula  comes  out  lower  than  0.30,  but  in  preliminary  cal- 


154 


THE  MAGNETIC  CIRCUIT 


[ART.  49 


ta? 


B 


eolations  it  is  advisable  to  use  at  least  0.30.  When  a  synchronous 
motor  is  working  near  the  limit  of  its  overload  capacity,  the  influ- 
ence  of  the  distorting  ampere-turns  is  particularly  important,  and 
in  estimating  the  overload  capacity  of  a  synchronous  motor  it  is 
better  to  be  on  the  safe  side  and  to  take  the  value  of  the  numerical 
coefficient  in  eq.  (81)  somewhat  higher  than  0.30.  The  value  of 
this  coefficient  varies  within  wider  limits  than  that  of  the  corre- 
sponding coefficient 
in  formula  (79)  ; 
but,  fortunately,  it 
affects  the  perform- 
ance to  a  lesser  de- 
gree (see  Art.  51). 

49.  The  Blondel 
Performance  Dia- 
gram of  a  Syn- 
chronous Machine 
with  Salient  Poles. 
Having  replaced  the 
actual  armature 
reaction  by  two 
m.m.fs.  Md  and  Mt 
(Fig.  39)  the  elec- 
tromagnetic rela- 
tions in  the  machine 
become  those  indi- 
cated in  Figs. 40  and 
41.  Fig.  40  refers 
to  a  generator  and 
is  analogous  to  Fig. 
37;  Fig.  41  refers 
to  a  motor  and  is 
analogous  to  Fig. 

38.  The  polygon  OABD,  which  represents  the  relation  between 
the  terminal  and  the  induced  voltages,  is  the  same  as  before,  but 
the  induced  voltage  E  is  now  considered  as  a  resultant  of  the 
voltages  En  and  Et  induced  by  the  real  and  the  fictitious  poles 
respectively.1  In  the  generator  the  fictitious  poles  lag  behind 

1  The  subscript  n  stands  for  net,  to  agree  with  the  m.m.f.  Mn  used  later 
on;  the  subscript  t  stands  for  transverse. 


FIG.  40. — The  performance  diagram  of  a  synchro- 
nous generator,  with  salient  poles. 


CHAP.  VIII]    REACTION  IN  SYNCHRONOUS  MACHINES 


155 


the  real  ones,  in  a  motor  they  lead  the  real  poles.  Hence,  in  the 
generator  diagram,  Et  lags  90  degrees  behind  En,  while  in  the 
motor  diagram  it  leads  En  by  90  degrees. 

In  the  case  of  a  generator  the  problem  usually  is  to  find  the 
field  excitation  Mf  neces- 
sary for  maintaining  a 
required  terminal  voltage 
e,  with  a  given  current  i 
and  at  a  given  power- 
factor  cos  <£.  First,  the 
figure  OABD  is  con- 
structed, or  else  the 
values  of  E  and  <f>'  are 
determined  from  eqs. 
(75),  (76),  and  (76o). 
In  order  to  find  the 
ampere-turns  required  on 
the  main  poles  it  is  neces- 
sary to  determine  the 
voltage  En  induced  by 
them.  For  this  purpose 
the  angle  /?  must  first  be 
known,  for 

En  =  E  cos  /?.  .     (82) 

As  an  intermediate  step, 
it  is  necessary  to  express 
Et  through  the  ampere- 
turns  Mt,  which  are  the 
cause  of  Et.  The  m.m.f . 
Mt  is  small  as  compared 
to  the  total  number  of 
ampere-turns  on  the  real 
poles;  hence,  the  lower 
straight  part  of  the  no- 
load  saturation  curve  of 

.  .  FIG.  41. — The  performance  diagram  of  a  syn- 

the  machine  can  be  used  chronous  ^^  with  salient  poles>    ' 

to  express   the  relation 

between  Mt   and   Et.      Let   v  be  the  voltage   corresponding  to 

one   ampere-turn    on  the  lower  part  of  the  no-load  saturation 


156  THE   MAGNETIC   CIRCUIT  [ART.  49 

curve;  then  Et  =Mtv.     Substituting  the  value  of  Mt  from  eq.  (81), 
we  have 


#,=#/  cos  (<£'+/?),        .....     (83) 
where 

(84) 


Et'  is  a  known  quantity  introduced  for  the  sake  of  brevity.  The 
angle  ^  in  formula  (83)  is  expressed  through  <j>'  and  ft  because, 
from  Fig.  40, 

^=<£'+/?  ........     (85) 

Another  relation  between  Et  and  /?  is  obtained  from  the  triangle 
ODG,  from  which 

Et  =  Esmp  ........     (86) 

A  comparison  of  eqs.  (83)  and  (86)  gives  that 
Et'  cos  (<£'  +fi)  =  E  sin  ft 
Expanding  and  dividing  throughout  by  cos  /?  we  find  the  relation 


(87) 


from  which  the  angle  /?  can  be  determined,  and  then  En  calculated 
byeq.  (82).  - 

The  next  step  is  to  take  from  the  no-load  saturation  curve  the 
value  Mn  of  the  net  excitation  necessary  on  the  main  poles  in  order 
to  induce  the  voltage  En.  The  real  excitation  M  /  must  be  larger, 
because  part  of  it  is  neutralized  by  the  direct  armature  raction  Mj. 
We  thus  have 

......     (88) 


where  Md  is  calculated  from  eq.  (79),  the  angle  ([>  being  known 
from  eq.  (85).  When  the  load  is  thrown  off,  the  only  excitation 
left  is  Mf,  let  it  correspond  to  a  voltage  e0  on  the  no-load  satura- 
tion curve.  From  e0  and  e  the  per  cent  voltage  regulation  of  the 
machine  is  determined  from  its  definition  as  the  ratio  (e0—e)/e. 

The  same  general  method  and  the  same  equations  apply  in  the 
case  of  Fig.  41,  when  one  is  required  to  determine  a  point  on  one  of 
the  phase  characteristics  of  a  synchronous  motor.  The  beginner 
must  be  careful  with  the  sign  minus  in  the  case  of  the  motor. 


CHAP.  VIII]    REACTION   IN  SYNCHRONOUS  MACHINES  157 

Since  <f>'  >  90  degrees,  the  angle  /?  and  the  voltage  Et  are  negative. 
The  angle  </>2  also  is  usually  negative.  The  cases  of  a  leading 
current  in  the  generator  and  of  a  lagging  current  in  the  motor  are 
obtained  by  assigning  the  proper  value  and  sign  to  the  angle  </>. 
For  the  application  of  the  Blondel  diagram  to  the  determination 
of  the  overload  capacity  of  a  synchronous  motor  see  the  reference 
given  near  the  end  of  Art.  47. 

A  synchronous  motor  is  sometimes  operated  at  no  load,  and  at 
such  a  value  of  the  field  current  that  the  machine  draws  reactive 
leading  kilovolt-amperes  from  the  Ikie,  thus  improving  the 
power-factor  of  the  system.  In  such  a  case  the  machine  is  called  a 
synchronous  condenser,  or  better,  a  phase  adjuster.  The  diagram  in 
Fig.  41  is  greatly  simplified  in  this  case  because  the  energy  com- 
ponent of  the  current  can  be  neglected,  as  well  as  the  drop  ir, 
and  the  e.m.f.  Et.  We  then  have  i=i2  =  id,  and  En=E=e+ix. 
The  direct  armature  reaction  is  determined  from  eq.  (79)  in  which 
^=90.  When  the  motor  is  underexcited  and  draws  a  lagging 
current  from  the  line,  i  is  to  be  considered  negative,  or  ^=270 
degrees.  The  same  simplified  diagram  applies  to  a  polyphase 
rotary  converter,  operated  from  the  alternating-current  side,  at  no 
load. 

Prob.  14.  It  is  required  to  calculate  the  field  current  and  per  cent 
voltage  regulation  of  a  12-pole,  150  kva.,  2300- volt,  60-cycle,  Y-connected 
alternator,  at  a  power  factor  of  85  per  cent  lagging.  The  machine  has 
two  slots  per  pole  per  phase,  and  is  provided  with  a  full-pitch  winding, 
the  number  of  turns  per  pole  per  phase  being  18.  The  armature  resist- 
ance per  phase  of  Y  is  0.67  ohm,  the  reactance  is  3.5  ohm.  The  number 
of  field  turns  per  pole  is  200.  The  no-load  saturation  curve  is  plotted 
for  the  line  voltage  (not  the  phase  voltage),  and  at  first  is  a  straight  line 
such  that  at  1800  volts  the  field  current  is  17.4  amp.  The  working  part 
of  the  no-load  saturation  curve  is  as  follows : 

Kilovolts 2.2        2.4    2.5    2.6    2.7    2.78 

Field  current,  amp 22          25      27      30      34       40 

Ans.  31  amp.;  14.3  per  cent. 

Prob.  15.  Show  that  in  the  foregoing  machine  the  short-circuit 
current  is  equal  to  about  two  and  a  half  times  the  rated  current,  at  the 
field  excitation  which  gives  the  rated  voltage  at  no-load.  Hint:  The 
short-circuit  curve  is  a  straight  line  so  that  one  can  first  calculate  the 
field  current  for  any  assumed  value  of  the  armature  current  and  e=0. 

Prob.  16.  From  the  results  of  the  calculations  of  the  preceding 
problem  show  that  the  cross-magnetizing  effect  and  the  ohmic  drop  are 
negligible  under  short-circuit,  in  the  machine  under  consideration. 


158  THE  MAGNETIC  CIRCUIT  [ART.  50 

Assuming  that  r  is  usually  small  as  compared  to  x,  describe  a  simple 
method  for  calculating  the  short-circuit  curve,  using  only  the  reactance 
of  the  machine  and  the  demagnetizing  ampere-turns  of  the  armature. 
In  practice,  the  influence  of  the  neglected  factors  is  accounted  for  in  short- 
circuit  calculations  by  taking  sin  (p  in  formula  (79)  as  equal  to  between 
0.95  and  0.98  instead  of  unity. 

Prob.  17.  Plot  the  no-load  phase  characteristic  of  the  machine 
specified  in  problem  14,  when  it  is  used  as  a  motor.  The  iron  loss  and 
friction  amount  to  8.5  kw. 

Ans.    Field  amperes 14.9        23.4        32.6 

Armature  amperes 30          2.13          30 

Prob.  18.  The  machine  specified  in  problem  14  is  to  be  used  as  a 
motor,  at  a  constant  input  of  150  kw.  Plot  its  phase  characteristics, 
i.e.,  the  curves  of  the  armature  current  and  of  power-factor  against  the 
field  current  as  abscissae. 

Ans.    Field  amperes 32.6        24.3          16.65 

Armature  amperes . .     47 . 00      37 . 65         47 . 00 
Power-factor 0 . 80        1 . 00  0 . 80 

Prob.  19.  Write  complete  instructions  for  the  predetermination  of 
the  regulation  of  alternators  and  of  the  phase  characteristics  of  synchro- 
nous motors,  by  BlondePs  method.  The  instructions  must  give  only 
the  successive  steps  in  the  calculations,  without  any  theory  or  explana- 
tions. Write  directions  and  formula?  on  the  left-hand  side  of  the  sheet, 
and  a  numerical  illustration  on  the  right-hand  side  opposite  it. 

Prob.  20.  Calculate  the  overload  capacities  of  the  foregoing  motor 
at  field  currents  of  25  amp.  and  35  amp.,  by  the  two  methods  described 
in  the  articles  refered  to  near  the  end  of  Art.  47. 

Prob.  21.  Show  that  for  a  machine  with  non-salient  poles  BlondePs 
and  Potier's  diagrams  are  identical. 

50.  The  Calculation  of  the  Value  of  the  Coefficient  of  Direct 
Reaction  in  Eq.  (79) ^  The  average  value  0.83  of  the  ratio  of  the 
effective  armature  m.m.f.  over  a  pole-face  to  the  maximum  m.m.f . 
at  the  center  of  the  pole  is  given  in  Art.  48  without  proof.  The 
following  computations  show  the  reasonable  theoretical  limits  of 
this  ratio.  If  the  armature  m.m.f.  (direct  reaction)  at  the  center 
of  the,  N  pole  (Fig.  39)  is  M,  its  value  at  some  other  point  along 
the  air-gap  is  M  cos  x,  where  x  is  measured  in  electrical  radians. 
Let  the  permeance  of  the  active  layer  of  the  machine  per  electrical 
radian  be  (P  at  the  center  of  the  pole,  and  let  this  permeance  vary 
along  the  periphery  of  the  armature  according  to  a  law  f(x) ,  so 
that  at  a  point  determined  by  the  abscissa  x  the  permeance  per 

1  This  and  the  next  article  can  be  omitted,  if  desired,  without  impairing 
the  continuity  of  treatment. 


CHAP.  VIII]    REACTION  IN  SYNCHRONOUS  MACHINES  159 

electrical  radian  is  (Pf(x).  The  function/Or)  must  be  periodic  and 
such  that  /(O)  =  1,  and  fQn)  =  0,  /(TT)  =  1,  etc.,  because  the  perme- 
ance reaches  its  maximum  value  under  the  centers  of  the  poles 
and  is  practically  nil  midway  between  the  poles. 

The  direct  armature  m.m.f.,  acting  alone,  without  any  excita- 
tion on  the  poles,  would  produce  in  each  half  of  a  pole  a  flux 

0=  C+**.Mcosx(Pf(x)dx. 
Jo 

The  magnetomotive  force  Md  placed  on  the  real  poles,  acting 
alone,  must  produce  the  same  total  flux,  so  that 


Equating  the  two  preceding  expressions  we  get 

MC**cosxf(x)dx~MdC**f(x)dx.      .    .     .     (89) 

«/0  »/0 

The  ratio  of  Md  to  M  can  be  calculated  from  this  equation,  by 
assuming  a  proper  law  f(x)  according  to  which  the  permeance  of 
the  active  layer  varies  with  x,  in  poles  of  the  usual  shapes.  Hav- 
ing a  drawing  of  the  armature  and  of  a  pole,  the  magnetic  field  can 
be  mapped  out  by  the  judgment  of  the  eye,  assisted  if  necessary 
by  Lehmann's  method  (Art.  41  above).  A  curve  can  then  be 
plotted,  giving  the  relative  permeances  per  unit  peripheral  length, 
against  x  as  abscissae.  Thus,  the  function  f(x)  is  given  graphically, 
and  the  two  integrals  which  enter  into  eq.  (89)  can  be  determined 
graphically  or  be  calculated  by  Simpson's  Rule.  Or  else, 
f(x)  can  be  expanded  into  a  Fourier  series  and  the  integration 
performed  analytically.  Such  calculations  performed  on  poles  of 
the  usual  proportions  give  values  of  Md/M  of  between  0.81  and 
0.85. 

It  is  also  possible  to  assume  for  f(x)  a  few  simple  analytical 
expressions,  and  integrate  eq.  (89)  directly.  Take  for  instance 
f(x)  =cos2  x.  By  plotting  this  function  against  x  as  absciss®  the 
reader  will  see  that  the  function  becomes  zero  midway  between 
the  poles,  is  equal  to  unity  opposite  the  centers  of  the  poles,  and 
has  a  reasonable  general  shape  at  intermediate  points.  Substi- 
tuting cos2  x  for  f(x)  into  eq.  (89)  and  integrating,  gives  f  M  = 
,  from  which  Md/M  =0.85. 


160  THE   MAGNETIC  CIRCUIT  [ART.  51 

Another  extreme  assumption  is  that  of  poles  without  chamfer, 
with  a  constant  air-gap.  Neglecting  the  fringe  at  the  pole-tips, 
f(x)  =  1  from  x  =0  to  x  =6,  and  f(x)  =0  from  x  =6  to  x  =%n.  Inte- 
grating eq.  (89)  between  the  limits  0  and  6  we  obtain 


(90) 


The  poles  usually  cover  between  60  and  70  per  cent  of  the  periph- 
ery. For  0=0.6(j7r)  the  preceding  equation  gives  Md/M=0.86, 
and  for  6=0.7(%n),  Md/M=0.81. 

Prob.  22.  Let  the  permeance  of  the  active  layer  decrease  from  the 
center  of  the  poles  according  to  the  straight-line  law,  so  that 

/(*)-!  -(2/*)s. 

What  is  the  ratio  of  Md/Mf  Ans.  0.81  1  . 

Prob.  23.  The  permeance  of  the  active  layer  decreases  according 
to  a  parabolic  law,  that  is,  as  the  square  of  the  distance  from  the  center 
of  the  poles.  What  is  the  ratio  of  Md/M?  Ans.  0.774. 

Prob.  24.  The  law  f(x}  =cos2  x  assumed  in  the  text  above  presup- 
poses that  the  permeance  varies  according  to  a  sine  law  of  double 
frequency  with  a  constant  term,  because  cos2  z=£  +£  cos  2x.  In  reality, 
the  permeance  varies  more  slowly  under  the  poles  and  more  rapidly 
between  the  poles  than  this  law  presupposes  (Fig.  39).  A  correction  can 
be  brought  in  by  adding  another  harmonic  of  twice  the  frequency  to  the 
foregoing  expression,  thus  making  it  un  symmetrical,  and  of  the  form 
f(x)  =a  +  b  cos  2x+c  cos  4x.  Show  that  f(x)  =2  cos2  x  —  cos4  x  contains 
the  largest  relative  amount  of  the  fourth  harmonic,  consistent  with  the 
physical  conditions  of  the  problem,  and  compare  graphically  this  curve 
with/(z)  =cos2  x. 

Prob.  25.  What  is  the  value  of  Md/M  for  the  form  of  f(x)  given  in 
the  preceding  problem?  Ans.  0.815. 

Prob.  26.  Plot  the  curve  /(x)  for  a  given  machine,  estimating  the 
permeances  by  Lehmann's  method,  and  determine  the  value  of  the  coef- 
ficient in  formula  (79). 

51.  The  Calculation  of  the  Value  of  the  Coefficient  of  Transverse 
Reaction  in  Eq.  (81).  The  average  value  0.33  of  the  ratio  of  the 
maximum  distorting  armature  m.m.f  .  to  the  equivalent  number  of 
ampere-turns,  Mt,  on  the  fictitious  poles  is  given  in  Art.  48  without 
proof.  The  following  computations  show  the  reasonable  theoret- 
ical limits  of  this  ratio.  The  problem  is  more  complicated  than 
that  of  finding  the  ratio  of  Md/M,  because  there  the  field  ampere- 
turns,  the  actual  demagnetizing  armature-m.m.f.,  and  the  equiva- 
lent ampere-turns  Md  are  all  acting  on  the  same  permeance  of  the 


CHAP.  VIII]     REACTION  IN  SYNCHRONOUS  MACHINES  161 

active  layer,  and  the  wave  form  of  the  flux  is  very  little  affected  by 
the  direct  armature  reaction.  In  the  case  of  the  transverse  reac- 
tion, however,  the  wave  form  of  the  flux  produced  by  the  actual 
cross-magnetizing  ampere-turns  of  the  armature  is  entirely  differ- 
ent from  that  produced  by  the  coil  Mt  acting  on  the  fictitious  pole 
(Fig.  39).  Namely,  the  actual  curve  of  the  transverse  flux  has  a 
large  "  saddle  "  in  the  middle,  due  to  the  large  reluctance  of  the 
space  between  the  real  poles.  The  flux  distribution  produced  by 
the  fictitious  poles  is  practically  the  same  as  that  under  the  main 
poles,  the  two  sets  of  poles  being  of  the  same  shape. 

The  addition  of  the  vectors  Et  and  En  in  Figs.  40  and  41  is  legit- 
imate only  when  Et  is  induced  by  a  flux  of  the  same  density  dis- 
tribution as  En,  and  this  is  the  reason  for  representing  the  trans- 
verse reaction  as  due  to  fictitious  poles  of  the  same  shape  as  the  real 
poles.  Therefore,  for  the  purposes  of  computation,  the  flux  dis- 
tribution, produced  by  the  actual  distorting  ampere-turns  on  the 
armature,  is  resolved  into  a  distribution  of  the  same  form  as  that 
produced  by  the  main  poles  and  into  higher  harmonics.  The  m.m.f . 
Mt  is  calculated  so  as  to  produce  the  first  distribution  only.  This 
fundamental  curve  is  not  sinusoidal,  but  will  have  a  shape  depend- 
ing on  the  shape  of  the  pole  shoes.  The  effect  of  the  sinusoidal 
higher  harmonics  on  the  value  of  Et  is  disregarded,  or  it  can  be 
taken  into  account  by  correcting  the  value  of  the  coefficient  in 
formula  (81)  from  the  results  of  tests. 

The  first  harmonic  of  the  armature  distortion  m.m.f.  is  M  sin  x, 
because  this  m.m.f.  reaches  its  maximum  between  the  real  poles; 
x  is  measured  as  before  from  the  centers  of  the  real  poles.  The 
permeance  of  the  active  layer,  with  reference  to  the  real  poles,  can 
be  represented  as  before  by  (Pf(x) .  The  flux  density  produced  by 
the  transverse  reaction  of  the  armature  at  a  point  denned  by  the 
abscissa  x  is  therefore  proportional  to  M  sin  x  (Pf(x) .  The  per- 
meance of  the  active  layer  with  reference  to  the  fictitious  poles  is 
<Pf(x+fa)-  The  flux  density  under  the  fictitious  poles  follows 
therefore  the  law  Mt(Pf(x+%7c).  As  is  explained  before,  the  two 
distributions  of  the  flux  density  differ  widely  from  one  another, 
and  the  real  distribution  is  resolved  into  the  fictitious  distribution, 
and  higher  sinusoidal  harmonics ;  the  prominent  third  harmonic  is 
clearly  seen  in  Fig.  39.  Thus,  we  have,  omitting  (P, 

M  sin  xf(x)  =Mtf(x+fa)  +A3  sin  3x+A5  sin  5z+etc.      (91) 


162  THE  MAGNETIC  CIRCUIT  [ABT.  51 

In  order  to  determine  Mt,  the  usual  method  is  to  mulitply  both 
sides  of  this  equation  by  sin  x  and  integrate  between  0  and  xf 
because  then  all  upper  harmonics  give  terms  equal  to  zero.  In 
this  particular  case  the  limits  of  integration  can  be  narrowed  down 
to  0  and  \K,  because  the  symmetry  of  the  curve  is  such  that  the 
segment  between  0  and  \n  is  similar  to  all  the  rest.  Thus,  we 
get 

M  f^sin2  xf(x)dx  =Mt  f^sin  xf(x+fr)dx.       .     (92) 
«/o  a/o 

From  this  equation  the  ratio  Mt/M  can  be  calculated  by  the 
methods  shown  in  Art.  50,  i.e.,  by  assuming  reasonable  forms  of 
the  function  f(x).  Taking  again  f(x)  =cos2  x  and  integrating  eq. 
(92)  we  get  &nM  =%Mt,  from  which  Mt/M  =0.295.  Taking  the 
other  extreme  case,  viz.,  f(x)  =1  from  x=0  to  x=6,  and  f(x)  =0 
from  x  =6  to  x  =%n,  gives,  after  integration 


(93) 


For  0=0.6(4*:),  M,/M=0.29;  for  0=0.7(4*),  M</M=0.39.1  It 
will  be  noted  that  the  cross-magnetizing  action  of  the  armature 
increases  considerably  with  the  increasing  ratio  of  pole-arc  to  pole- 
pitch,  while  the  direct  reaction  slowly  diminishes  with  the  increase 
of  this  ratio.  In  machines  intended  primarily  for  lighting  pur- 
poses it  is  advisable  to  use  a  rather  small  ratio  of  pole-arc  to  pole- 
pitch,  in  order  to  reduce  transverse  reaction  which  affects  the  volt- 
age regulation  at  high  values  of  power-factor  in  particular. 

Prob.  27.  What  is  the  value  of  Mt/M  for  the  form  of  f(x)  given  in 
problem  24;  namely,  for/(z)  =  2  cos2  z-cos4  x?  Ans.  0.368. 

Prob.  28.  Determine  the  numerical  value  of  the  coefficient  in  formula 
(81)  for  the  machine  used  in  problem  26. 

1  These  values  are  higher  than  those  derived  by  E.  Arnold.  The  fact 
that  Arnold's  values  for  the  coefficient  of  transversal  reaction  are  low  has 
been  pointed  out  by  Sumec  in  Elektrotechnik  und  Maschinenbau,  1906,  p. 
67;  also  by  J.  A.  Schouten,  in  his  article  "  Ueber  den  Spannungsabfall  mehr- 
phasiger  synchroner  Maschinen,  "  Elektrotechnische  Zeitschrift,  Vol.  31  (1910), 
p.  877. 


CHAPTER  IX 

ARMATURE    REACTION    IN    DIRECT-CURRENT 
MACHINES 

52.  The   Direct   and   Transverse   Armature   Reactions.     Let 

Fig.  42  represent  the  developed  cross-section  of  a  part  of  a  direct- 
current  machine,  either  a  generator  or  a  motor.  For  the  sake  of 
simplicity  the  brushes  are  shown  making  contact  directly  with  the 
armature  conductors,  omitting  the  commutator.  Electrically 
this  is  equvialent  to  the  actual  conditions,  because  the  commutator 
segments  are  soldered  to  the  end-connections  of  the  same  conduc- 
tors. The  brushes  are  shifted  by  a  distance  d  from  the  geometrical 
neutral,  to  insure  a  satisfactory  commutation;  d  being  expressed 
hi  centimeters,  measured  along  the  armature  periphery,  the  same 
as  the  pole-pitch  T. 

The  actual  armature  conductors  and  currents  are  replaced,  for 
each  pole-pitch,  by  a  current  sheet,  or  belt,  of  the  same  strength. 
Let,  for  instance,  the  pole-pitch  be  40  cm.,  and  let  the  machine 
have  120  armature  conductors  per  pole.  If  the  current  per  con- 
ductor is  100  amp.,  the  total  number  of  ampere-conductors  per 
pole  is  12,000 ;  the  total  current  of  the  equivalent  belt,  which  con- 
sists of  one  wide  conductor,  must  be  12,000  amp.,  or  300  amp.  per 
linear  cm.  of  the  pole-pitch.  The  latter  value,  or  the  number  of 
armature  ampere-conductors  per  centimeter  of  periphery,  is  some- 
times called  the  specific  electric  loading  of  the  machine.  The  mag- 
netic action  of  the  equvialent  current  sheet  on  the  magnetic  flux  of 
the  machine  is  practically  the  same  as  that  of  the  actual  armature 
conductors,  because  in  a  direct-current  machine  the  slots  are  com- 
paratively numerous  and  small.  The  current  in  the  cross-hatched 
belts  is  supposed  to  flow  from  the  reader  into  the  paper,  and  the  cur- 
rent in  the  belts  marked  with  dots — toward  the  reader.  With  the 
directions  of  the  flux  and  of  the  current  shown  in  the  figure,  the 
directions  of  rotation  of  the  machine  when  working  as  a  generator 
and  as  a  motor  are  those  shown  by  the  arrow-heads  (see  Art.  24). 

163 


164 


THE  MAGNETIC  CIRCUIT 


[ART.  52 


The  polarity  of  the  brushes  cannot  be  indicated  without  knowing 
the  actual  connections  in  the  winding.  It  is  preferable,  therefore, 
for  our  purposes  to  designate  the  brushes  as  E  and  W  (east  and 
west),  according  to  their  position  with  respect  to  the  poles  of  the 
machine,  the  observer  looking  from  the  commutator  side.  The 
whole  interpolar  regions  to  the  right  of  the  north  poles  can  be  called 
the  eastern  regions,  those  to  the  left  the  western  regions;  the  same 
notation  can  be  also  applied  to  the  commutating  poles. 

The  armature  currents  exert  a  two-fold  action  upon  the  main 
field  of  the  machine:  they  partly  distort  it,  and  partly  weaken  it. 
For  the  purposes  of  theory  and  calculation  it  is  convenient  to 
separate  these  two  actions,  the  same  as  in  the  case  of  the  synchro- 


cj 


0,1 


FIG.   42. — The   direct   and   transverse    armature    reactions  in   a 
direct-current   machine. 

nous  machine  in  the  preceding  chapter.  Let  the  sheets  of  current 
be  divided  into  parts  denoted  by  the  letters  D  and  T  with  sub- 
scripts corresponding  to  their  location  with  reference  to  the  poles 
and  brushes.  The  belts  denoted  by  D  are  comprised  within  the 
space  d,  one  each  side  of  the  geometrical  neutrals;  those  denoted 
by  T  are  (Jr— <J)  centimeters  wide. 

The  belts  D  exert  a  direct  demagnetizing  action  upon  the  poles. 
Namely,  the  belts  Dn  Dn  can  be  considered  as  two  sides  of  a  coil  the 
axis  of  which  is  along  the  center  line  CnCn'.  The  m.m.f.  of  this 
coil  opposes  that  of  the  field  coil  on  the  north  pole.  In  the  same 
way,  the  m.m.f.  of  the  coil  DSDS  opposes' the  action  of  the  field  coil 
on  the  south  pole.  The  foregoing  is  true  no  matter  what  the 
actual  connections  of  the  armature  conductors  are,  provided  that 
the  winding-pitch  is  nearly  100  per  cent.  With  a  fractional-pitch 


CHAP.  IX]    ARMATURE  REACTION  IN  D.C.  MACHINES  165 

winding  the  currents  within  each  D  belt  flow  partly  in  the  opposite 
directions  and  neutralize  each  other's  action. 

Let  the  specific  electric  loading  of  the  machine,  as  defined 
above,  be  (AC) .  Then,  with  a  full-pitch  winding,  the  demagnetiz- 
ing ampere-turns  per  pole  are1 

M,=(AC)d (94) 

The  belts  TeTe  constitute  together  a  coil  the  center  of  which  is 
along  the  axis  OeOer]  the  adjacent  belts  TwTwiono.  a  coil  with  its 
axis  along  0W0W'.  The  m.m.f.  distribution  of  these  coils  is  indi- 
cated by  the  broken  line  ABC,  which  shows  that  the  T  belts  pro- 
duce a  transverse  armature  reaction.  The  line  ABC  is  also  the 
curve  of  the  flux  density  distribution  which  would  be  produced  by 
the  transversal  reaction  alone,  if  the  active  layer  of  the  machine 
were  the  same  throughout  (non-salient  poles).  On  account  of  a 
much  higher  reluctance  of  the  paths  in  the  interpolar  regions  the 
flux  density  there  is  much  lower,  and  is  shown  by  the  dotted  lines. 
The  actual  distribution  of  the  field  in  a  loaded  machine  is  obtained 
considering  from  point  to  point  the  field  and  armature  m.m.fs. 
acting  upon  the  individual  magnetic  paths. 

The  transverse  reaction  opposes  the  field  m.m.f.  under  one-half 
of  each  pole  and  assists  it  under  the  other  half,  so  that  the  main 
field  is  distorted.  In  a  generator  the  field  is  shifted  in  the  direc- 
tion of  rotation,  in  a  motor  it  is  crowded  against  the  direction  of 
rotation  of  the  armature.  This  is  the  same  as  what  happens  in 
synchronous  machines,  when  the  armature  is  revolving  and  the 
poles  stationary  (see  Fig.  36). 

The  brushes  must  be  shifted  in  the  same  direction  in  which  the 
flux  is  shifted,  because  the  magnetic  neutral  is  displaced  with 
respect  to  the  geometric  neutral.  Usually,  the  brushes  are  shifted 
beyond  the  magnetic  neutral,  in  order  to  obtain  a  proper  flux  den- 
sity for  commutation.  Namely,  to  assist  the  reversal  of  the  cur- 
rent in  the  conductors  which  are  short-circuited  by  the  brushes, 
these  conductors  must  be  brought  into  the  fringe  of  a  field  of  such 
a  direction  as  assists  the  commutation.  In  the  case  of  a  gener- 
ator this  means  the  field  under  the  influence  of  which  the  conduc- 

1  The  effect  of  the  coils  short-circuited  under  the  brushes  is  not  considered 
separately,  for  the  sake  of  simplicity.  For  an  analysis  of  the  reaction  of  the 
short-circuited  coils  upon  the  field  see  E.  Arnold,  Die  Gleichstrommaschine 
Vol.  1  (1906),  Chap.  23. 


166  THE  MAGNETIC  CIRCUIT  [AET.  53 

tors  come  after  the  commutation.  In  a  motor  the  armature  cur- 
rent flows  against  the  induced  e.m.f.;  it  is  therefore  the  field  which 
cuts  the  conductors  before  the  commutation  that  assists  the  rever- 
sal of  the  current.  This  explains  the  direction  of  the  shift  of  the 
brushes  in  the  two  cases.  The  student  should  make  this  clear  to 
himself  by  considering  in  detail  the  directions  of  the  currents  and 
of  the  induced  voltages  in  a  particular  case. 

The  maximum  m.m.f.  per  pole  produced  by  the  distorting 
belts  is  equal  to  (AC)  (%T—d),  but  since  this  m.m.f.  acts  along  the 
interpolar  space  of  high  reluctance  its  effect  is  not  large  (except  in 
machines  with  commutating  poles).  Of  much  more  importance 
is  the  action  of  the  distorting  belts  under  the  main  poles.  At  each 
pole-tip  the  armature  m.m.f.  is 

w,       (95) 


where-  w  is  the  width  of  the  pole  shoe.  This  m.m.f.  decreases 
according  to  the  straight  line  law  to  the  center  of  each  pole  and 
is  of  opposite  signs  at  the  two  tips  of  the  same  pole. 

Prob.  1.  Determine  the  polarity  of  the  brushes  in  Fig.  42  for  a 
progressive  and  a  retrogressive  winding,  in  the  case  of  a  generator  and 
of  a  motor. 

Prob.  2.  Indicate  the  D  and  the  T  belts  in  a  fractional-pitch  winding 
(a)  with  the  brushes  in  the  geometric  neutral,  and  (b)  with  the  brushes 
shifted  by  d. 

Prob.  3.  A  500  kw.,  230  volt,  10  pole,  direct-current  machine  has 
a  full-pitch  multiple  winding  placed  in  165  slots.  There  are  8  con- 
ductors per  slot,  and  two  turns  per  commutator  segment.  What  are 
the  demagnetizing  ampere-turns  per  pole  when  the  brushes  are  shifted 
by  4  commutator  segments?  Ans.  3470. 

Prob.  4.  What  is  the  amplitude  of  the  broken  line  ABC  in  the 
preceding  machine?  Ans.  10850  amp-turns. 

Prob.  5.  For  a  given  machine,  draw  the  curves  of  the  flux  density 
distribution  under  a  pole,  at  no-load  and  at  full  load,  by  considering 
the  m.m.fs.  acting  upon  the  individual  paths,  and  the  reluctance  of  the 
paths.1 

53.  The  Calculation  of  the  Field  Ampere-turns  in  a  Direct- 
current  Machine  under  Load.  The  net  ampere-turns  per  pole  are 
determined  from  the  no-load  saturation  curve  of  the  machine  for 

1  For  details  and  examples  of  such  curves  see  Pichelmayer,  Dynamobau 
(1908),  p.  176;  Parshalland  Hobart,  Electric  Machine  Design  (1906),  p.  159; 
Arnold,  Die  Gleichstrommachine,  Vol.  1  (1906),  p.  324. 


CHAP.  IX]    ARMATURE  REACTION  IN  D.C.  MACHINES  167 

the  necessary  induced  voltage.  In  a  generator  the  induced  voltage 
is  equal  to  the  specified  terminal  voltage  plus  the  internal  ir  drop 
in  the  machine.  In  a  motor  the  induced  voltage  is  less  than  the 
line  voltage  by  the  amount  of  the  internal  voltage  drop.  The 
actual  ampere-turns  to  be  provided  on  the  field  poles  are  larger 
than  the  net  excitation  by  the  amount  necessary  for  the  compensa- 
tion of  the  armature  reaction. 

The  direct  reaction  is  compensated  for  by  adding  to  each  field 
coil  the  ampere-turns  given  by  eq.  (94).  For  instance,  in 
prob.  3  above,  3470  ampere-turns  per  pole  must  be  added  to  the 
required  net  excitation,  in  order  to  compensate  for  the  effect  of  the 
direct  armature  reaction.  The  necessary  shift  of  the  brushes  is 
only  roughly  estimated  from  one's  experience  with  previously 
built  machines,  though  it  could  be  determined  more  accurately 
from  the  distribution  of  flux  density  in  the  pole-tip  fringe.  The 
poles  usually  cover  not  over  70  per  cent  of  the  armature  periphery, 
so  that  the  distance  between  the  geometric  neutral  and  the  pole- 
tip  is  about  15  per  cent  of  the  pole  pitch.  In  preliminary  esti- 
mates, the  brush  shift  may  be  taken  to  be  about  10  per  cent  of  the 
pole  pitch;  this  brings  the  brushes  not  quite  to  the  pole-tips 
though  well  within  their  fringe.  In  actual  operation  a  smaller 
shift  may  be  expected.  In  machines  provided  with  commutating 
poles,  and  in  motors  intended  for  rotation  in  both  directions,  the 
brushes  are  usually  in  the  geometric  neutral,  so  that  the  demag- 
netizing action  is  zero. 

In  a  machine  with  a  low  saturation  in  the  teeth  and  in  the  pole- 
tips,  the  cross-magnetizing  m.m.f.  of  the  armature  does  not  affect 
the  magnitude  of  the  total  flux  per  pole,  because  the  flux  is 
increased  on  one-half  of  the  pole  as  much  as  it  is  reduced  on  the 
other  half.  It  is  shown  in  Art.  31  that  the  induced  e.m.f.  of  a 
direct-current  machine  is  independent  of  the  flux  distribution, 
provided  that  the  total  flux  is  the  same,  so  that  no  extra  field 
ampere-turns  are  necessary  in  such  a  machine  to  compensate  for 
the  distortion  of  the  flux. 

However,  the  teeth  and  the  pole-tips  are  usually  saturated  to 
a  considerable  extent,  so  that  the  flux  is  increased  on  one  side  of 
the  pole  less  than  it  is  reduced  on  the  other  side.  Thus,  the  useful 
flux  is  not  only  distorted  by  the  transverse  armature  reaction,  but 
is  also  weakened.  This  latter  effect  has  to  be  counterbalanced  by 
additional  ampere-turns  on  the  field  poles.  In  most  cases  these 


168 


THE  MAGNETIC  CIRCUIT 


[ART.  53 


additional  ampere-turns  are  estimated  empirically,  on  the  basis 
of  one's  previous  experience,  because  the  amount  is  not  large,  and 
a  correct  computation  is  rather  tedious.1 

The  theoretical  relation  between  the  distorting  ampere-turns 
and  the  field  ampere-turns  required  for  their  compensation  is  shown 
in  Fig.  43.  Let  OX  represent  the  no-load  saturation  curve  of  the 
machine  for  its  active  layer  only,  that  is,  for  the  air-gap,  the  teeth, 
and  the  pole  shoe,  if  the  latter  is  sufficiently  saturated.  The 
ordinates  represent  the  induced  e.m.f .  between  the  brushes,  or,  to 
another  scale,  the  useful  flux  per  pole ;  the  abscissae  give  the  corre- 
sponding values  of  the  field  ampere-turns  per  pole,  disregarding 
the  reluctance  of  the  field  poles,  field  yoke,  and  armature  core. 


a     a'  d     d1 

Ampere-turns  and  pole  face  areas 

FIG.  43.— A  construction  for  determining  the  field   m.m.f.  needed  for  the 
compensation  of  the  transverse  reaction. 

In  other  words,  the  abscissae  give  the  values  of  the  difference  of 
magnetic  potential  across  the  active  layer  of  the  machine,  at  no 
load.  It  is  proper  to  consider  here  the  m.m.fs.  across  the  active 
layer  only,  because  the  distorting  action  of  the  armature  extends 
only  over  this  layer.  No  matter  how  irregular  the  flux  distribu- 
tion in  the  air-gap  and  in  the  teeth  may  be,  the  flux  density  in  the 
pole  cores  and  in  the  yoke  is  practically  uniform  (compare  Fig.  36) . 

1  For  Hobart's  empirical  curves  for  estimating  the  field  excitation  required 
for  overcoming  the  armature  distortion  see  the  Standard  Handbook,  under 
'  Generators,  direct-current,  ampere-turns,  estimate." 


CHAP.  IX]    ARMATURE  REACTION   IN   D.C.   MACHINES  169 

For  the  sake  of  simplicity,  we  replace  the  actual  pole  shoe  by 
an  equivalent  one,  without  chamfer,  of  rectangular  shape,  and 
with  a  negligible  pole-tip  fringing.  The  ordinates  of  the  curve  OX 
represent  to  another  scale  the  average  flux  density  on  the  surface 
of  the  pole  shoe,  because  this  density  is  proportional  to  the  total 
useful  flux,  or  to  the  induced  voltage.  Let  ab  be  the  flux  density 
corresponding  to  the  required  induced  e.m.f.,  and  let  Oa=M  be 
the  necessary  net  m.m.f .,  without  the  transverse  reaction.  Let  ac 
and  ad  represent  the  distorting  ampere-turns,  M%,  at  the  pole-tips, 
as  given  by  eq.  (95) ;  then  Oc  and  Od  are  the  resultant  m.m.f s.  at 
the  pole-tips.  The  flux  density  at  the  pole-tips  of  the  loaded 
machine  is  then  equal  to  eg  and  dh  respectively. 

Since  the  distorting  ampere-turns  vary  directly  as  the  distance 
from  the  center  of  the  pole,  and  the  area  of  strips  of  equal  width 
is  the  same,  the  abscissae  from  a  represent  to  scale  either  distances 
along  the  pole  face,  or  areas  on  the  pole  face,  measured  from  the 
center  of  the  pole.  Thus,  the  part  gh  of  the  curve  OX  represents 
also  the  distribution  of  the  flux  density  under  the  pole,  in  the 
loaded  machine.  Likewise,  the  line  ef  represents  the  distribution 
of  the  flux  density  under  the  pole  at  no-load. 

Since  the  ordinates  represent  to  scale  the  flux  densities  and  the 
abscissae  the  areas  of  the  different  parts  under  the  poles,  the  area 
under  the  flux  density  distribution  curve  also  represents  to  scale 
the  total  flux  per  pole.  The  total  flux  at  no  load  is  represented  by 
the  area  of  the  reactangle  cefd,  and  to  the  same  scale,  the  flux  in  the 
loaded  machine  is  represented  by  the  area  cghd.  If  the  saturation 
curve  were  a  straight  line,  the  two  areas  would  be  equal,  so  that 
the  distortion  would  not  modify  the  value  of  the  total  flux  per  pole. 
In  reality,  the  area  geb  is  larger  than  the  area  bhf,  and  the  differ- 
ence between  the  two  represents  the  reduction  in  the  flux,  due  to 
the  transverse  armature  reaction. 

Let  now  the  field  excitation  be  increased  by  an  unknown  amount 
aaf  to  the  value  Oa'=M',  in  order  to  compensate  for  the  above 
explained  decrease  in  the  flux.  All  the  points  in  Fig.  43  are  shifted 
by  the  same  amount,  and  are  denoted  by  the  same  letters  with  the 
sign  "  prime."  The  new  flux  in  the  loaded  machine  is  represented 
by  the  area  c'g'h'd',  the  point  a'  being  the  center  of  the  line  c'd'  =cd. 
If  the  point  a!  has  been  properly  selected  we  must  have  the  con- 
dition that  the 

area  cefd=  area  c'g'h'd', (96) 


170  THE  MAGNETIC  CIRCUIT  [ART.  53 

that  is,  the  flux  corresponding  to  the  excitation  Oa  at  no-load  is 
the  same  as  the  flux  corresponding  to  the  excitation  Oa'  when  the 
machine  is  loaded.  The  problem  is  then,  knowing  the  point  a  and 
the  distance  cd=c'd'  to  find  the  point  a'  such  that  equation  (96) 
is  satisfied.  The  two  areas  have  the  common  part  c'g'bfd,  and 
the  parts  cee'c'  and  dff'd'  are  equal  to  each  other.  Therefore, 
eq.  (96)  is  satisfied  if  the 

area  be'g' =-  area  bfh' (97) 

The  position  of  the  point  a'  is  found  either  by  trials,  or  as  the 
intersection  of  the  curves  rrf  and  ss';  the  ordinates  of  the  curve  rrf 
represent  the  area  be'g'  for  various  assumed  positions  of  the  point 
a',  and  the  ordinates  of  the  curve  ssf  represent  the  corresponding 
values  of  the  area  bf'h'. 

Thus,  the  total  field  ampere-turns  required  for  a  direct-current 
generator  under  load  are  found  as  follows:  (a)  To  the  specified 
terminal  voltage  add  the  voltage  drop  in  the  armature,  under  the 
brushes,  and  in  the  windings  (if  any)  which  are  in  series  with  the 
armature,  viz.;  the  series  field  winding,  the  compensating  winding, 
and  the  interpole  winding.  This  will  give  the  induced  voltage  E. 
(b)  From  the  no-load  saturation  curve  find  the  excitation  corre- 
sponding to  E.  (c)  Estimate  the  amount  of  the  brush  shift  (if 
any)  and  calculate  the  corresponding  demagnetizing  ampere-turns 
according  to  eq.  (94) .  (d)  Determine  the  ampere-turns  aar  (Fig.  43) 
required  for  the  compensation  of  the  armature  distortion,  (e) 
Add  the  ampere-turns  calculated  under  (b),  (c)  and  (d).  In  the 
case  of  a  motor  subtract  the  voltage  drop  in  the  machine  from 
the  terminal  voltage,  to  find  the  induced  e.m.f.  E,  but  otherwise 
proceed  as  before. 

If  the  machine  is  shunt-wound  or  series-wound,  the  field  wind- 
ing is  designed  so  as  to  provide  the  necessary  maximum  number  of 
ampere-turns  at  a  required  margin  in  the  field  rheostat  for  the 
specified  voltage  or  speed  variations.  When  the  machine  is  com- 
pound wound,  the  shunt  winding  alone  must  supply  the  required 
number  of  ampere-turns  at  no-load.  The  series  ampere-turns  is, 
then,  the  difference  between  the  total  m.m.f.  required  at  full-load 
and  that  supplied  by  the  shunt-field.  When  a  generator  is  over- 
compounded,  the  shunt  excitation  is  larger  at  full  load  than  it  is  at 
no  load,  and  allowance  must  be  made  for  this  fact. 


CHAP.  IX]    ARMATURE  REACTION  IN  D.C.   MACHINES  171 

In  the  case  of  a  variable-speed  motor,  the  problem  of  predict- 
ing the  exact  speed  at  a  given  load  can  be  solved  by  successive 
approximations  only.  First,  this  speed  is  determined  neglecting 
the  transverse  armature  reaction  altogether,  or  assigning  to  it  a 
reasonable  value.  Then,  the  construction  shown  in  Fig.  43  is 
performed  for  a  few  speeds  near  the  approximate  value,  and  thus 
a  more  correct  value  of  the  speed  is  found  by  trials.  Or  else 
different  values  of  speed  are  assumed  first,  and  the  corresponding 
values  of  the  armature  current  are  found  for  each  speed.  These 
armature  currents  must  be  such  that,  considering  the  total  arma- 
ture' reaction,  the  field  m.m.f.  is  just  sufficient  to  produce  the 
required  counter-e.m.f.  in  the  armature.  The  details  of  the  solu- 
tion are  left  to  the  student  to  investigate.  He  must  clearly 
understand  that  the  problem  can  be  solved  only  for  a  given  or  an 
assumed  speed,  because  of  the  necessity  of  using  the  active  layer 
characteristic  OX  (Fig.  43). 

Prob.  6.  In  a  direct-current  machine  it  is  desired  to  have  a  full 
load  a  flux  density  of  not  less  than  3500  maxwells  per  sq.cm.  at  the 
pole-tip  at  which  the  commutation  takes  place;  the  m.m.f.  across  the 
active  layer  is  7500  amp  .-turns,  the  air-gap  is  11  mm.,  the  air-gap 
factor  1.25.  The  ratio  of  the  ideal  pole  width  to  the  pole  pitch  is  0.7. 
What  are  the  permissible  ampere-turns  on  the  armature,  per  pole? 
Solution:  The  net  m.m.f.  across  the  active  layers  at  the  pole-tip  under 
consideration  is  3500X0.8X1.1X1.25  =  3850  amp  .-turns.  Hence 
M2  =  %(AC)  X0.7r  =  7500 -3850 =3650.  Ans.  i(A<7)r=5200. 

Prob.  7.  The  specific  electric  loading  in  a  direct-current  machine 
is  250  amp.-conductors  per  centimeter;  the  average  flux  density  on 
the  pole-face  is  8.5  kl.  per  sq.  cm.,  at  no  load  and  at  the  rated  full  load 
terminal  voltage;  the  width  of  the  equivalent  ideal  pole  is  32  cm.  The 
estimated  internal  voltage  drop  at  full  load  is  about  5  per  cent  of  the 
terminal  voltage.  Calculate  the  ampere-turns  per  pole  required  to 
compensate  for  the  transverse  armature  reaction;  the  active  layer 
characteristic  (Fig.  43)  is  as  follows : 

M=    4        5        6        7        8        9        11       13    kilo  ampere-turns; 
5  =  5.40  6.75  7.75  8.50  8.90  9.20  9.55  9.78  kl.  per  sq.  cm. 

Ans.  950. 

Prob.  8.  For  the  preceding  machine,  calculate  the  total  required 
excitation  at  full  load,  per  pole;  the  brush  shift  is  8  per  cent  of  the 
pole-pitch;  the  pole-pitch  r  =  46  cm.;  2000  ampere-turns  are  required 
for  the  parts  of  the  machine  outside  the  active  layer. 

Ans.  12,000  amp  .-turns. 

Prob.  9.  A  shunt-wound  motor  is  designed  so  as  to  operate  at  a 
certain  speed  at  full  load.  Show  how  to  predict  its  speed  at  no-load. 


172  THE   MAGNETIC   CIRCUIT  [ART.  54 

Prob.  10.  Show  how,  in  a  compound-wound  motor,  the  total 
required  field  excitation  must  be  divided  between  the  shunt  and  series 
windings  in  order  to  obtain  a  prescribed  speed  regulation  between  no- 
load  and  full  load. 

Prob.  11.  Assume  the  curve  OX  in  Fig.  43  to  be  given  in  the  form 
of  an  analytic  equation,  B=f(M).  Show  that  the  unknown  excitation 
Oa'=M'  is  determined  by  the  equation  2M2f(M)  =  F(M'+M2)  - 
F(M'-M2),  where  the  function  F  is  such  that  dF(M)/dM=f(M}; 
MQ  =  Oa  is  the  excitation  corresponding  to  the  given  value  of  e  or  B  =  ab. 

Prob.  12.  Apply  the  formula  of  the  preceding  problem  to  the  case 
when  the  active  layer  characteristic  can  be  represented  by  (a)  the  log- 
arithmic curve  y  =  a log  (1  +bx) ;  (6)  the  hyperbola  y=gx/(h+x) ;  (c)  a 
part  of  the  parabola  (y*-y02)  =2p(x-Xo),  continued  as  a  tangent  straight 
line  passing  through  the  origin. 

54.  Commutating  Poles  and  Compensating  Windings.      The 

two  limiting  factors  in  proportioning  a  direct-current  machine  are, 
first,  the  sparking  under  the  brushes,  and  secondly,  the  armature 
reaction.  In  order  to  reverse  a  considerable  armature  current 
in  a  coil  during  the  short  interval  of  time  that  the  coil  is  under  a 
brush,  an  external  field  of  a  proper  direction  and  magnitude  is 
necessary.  In  ordinary  machines  (Fig.  42)  this  field  is  obtained 
by  shifting  the  brushes  so  as  to  bring  the  short-circuited  armature 
conductors  under  a  pole  fringe.  However,  with  this  method  the 
specific  electric  loading  and  the  armature  ampere-turns  must  be 
kept  below  a  certain  limit  with  reference  to  the  ampere-turns  on 
the  field;  otherwise  the  armature  reaction  would  weaken  the  field 
to  such  an  extent  as  to  reduce  the  flux  density  in  the  fringe  below 
the  required  value.  Therefore,  in  many  modern  machines,  instead 
of  moving  the  brushes  to  the  poles,  part  of  the  poles,  so  to  say,  are 
brought  to  the  brushes  (Fig.  44).  These  additional  poles  are 
called  commutating  poles  or  interpoles.  Their  polarity  is  under- 
stood with  reference  to  Fig.  42 :  Since  the  E  brush  had  to  be  shifted 
toward  the  north  pole,  now  a  north  interpole  is  placed  over  each  E 
brush. 

The  armature  belts  Te,  Te,  create  a  strong  m.m.f.  along  the  axis 
of  the  commutating  pole  Nc,  in  the  wrong  direction.  Therefore, 
the  winding  on  each  interpole  must  be  provided  first  with  a  num- 
ber of  ampere-turns  equal  and  opposite  to  that  of  the  armature, 
and  secondly  with  enough  additional  ampere-turns  to  establish 
the  required  commutating  flux.  These  additional  ampere-turns 
are  calculated  only  for  the  air-gap,  armature  teeth,  and  the  pole 
body  itself.  The  m.m.f.  required  for  the  armature  core  and  the 


CHAP.  IX]    ARMATURE  REACTION  IN  B.C.  MACHINES 


173 


yoke  of  the  machine  is  negligible,  because  the  commutating  flux  is 
small  as  compared  with  the  main  flux  and  is  displaced  with  respect 
to  it  by  ninety  electrical  degrees.  The  winding  on  the  interpoles 
is  connected  in  series  with  the  main  circuit  of  the  machine,  because 
the  armature  m.m.f.  is  proportional  to  the  armature  current,  and 
also  because  the  density  of  the  reversing  field  must  be  proportional 
to  the  current  undergoing  commutation. 

The  flux  density  under  the  interpoles  is  determined  from  the 
condition  that  the  e.m.f.  induced  in  the  armature  conductors  by 
the  commutating  flux  be  equal  and  opposite  to  the  e.m.f.  due  to 
the  inductance  of  the  armature  coils  undergoing  commutation. 
For  practical  purposes,  the  inductance  can  be  only  roughly  esti- 
mated (see  Art.  68  below),  but  on  the  other  hand  an  accurate  cal- 

Compensating         Commutating 
Main  pole        Winding  /pole  or  interpole 


Generator 


t 


Motor 


FIG.  44. — Interpoles  and  a  compensating  winding  in  a  direct-current  machine. 

culation  is  not  necessary,  because  the  number  of  ampere-turns  on 
the  interpole  is  easily  adjusted  by  a  shunt  around  its  winding,  as 
in  the  case  of  a  series  winding  on  the  main  poles.  Or  else  the 
commutating  flux  can  be  increased  by  "  shimming  up "  the 
interpoles.  The  m.m.f.  required  for  establishing  a  required 
commutating  flux  is  calculated  in  the  same  manner  as  in  the  case 
of  the  main  flux,  viz.,  the  saturation  curves  (Figs.  2  and  3)  are 
used  for  the  pole  body  and  the  teeth,  while  the  reluctance  of  the 
air-gap  is  estimated  as  is  explained  in  Arts.  36  and  37.  The 
commutating  poles  must  be  of  such  a  width  that  all  the  coils 
undergoing  commutation  are  under  their  influence. 

A  comparatively  large  leakage  factor,  say  between  1.4  and  1.5, 
or  over,  is  usually  assumed  for  the  commutating  poles,  on  account 
of  the  proximity  of  the  main  poles.  It  is  advisable  to  concentrate 
the  winding  near  the  tip  of  the  interpole,  in  order  to  reduce  the 


174  THE  MAGNETIC  CIRCUIT  [ART.  54 

magnetic  leakage.  The  leakage  factor  of  the  main  poles  is  also 
somewhat  increased  by  the  presence  of  the  interpoles;  this  is  one 
of  their  disadvantages.  Some  other  disadvantages  are :  the  ven- 
tilation of  the  field  coils  is  more  difficult,  and  a  smaller  ratio  of 
pole  arc  to  pole  pitch  must  be  used.  However,  the  advantages 
gained  by  the  use  of  commutating  poles  are  such  that  their  use 
is  rapidly  becoming  universal. 

The  interpole  winding  removes  the  effect  of  the  transverse 
belts  T,  T,  in  the  commutating  zone,  but  does  not  neutralize  their 
distorting  effect  under  the  main  poles.  Hence,  the  distortion  and 
its  accompanying  reduction  of  the  main  flux  are  practically  the 
same  as  without  the.  interpoles.  To  remove  this  distortion  a  com- 
pensating winding  (Fig.  44),  connected  in  series  with  the  main  cir- 
cuit of  the  machine,  is  sometimes  placed  on  the  main  poles.  The 
connections  are  such  that  the  compensating  winding  opposes  the 
magnetizing  action  of  the  armature  winding.  By  properly  select- 
ing the  specific  electric  loading  of  the  compensating  winding  the 
transverse  armature  reaction  under  the  poles  can  be  removed, 
either  completely  or  in  part.  This  winding  was  invented  inde- 
pendently by  Deri  in  Europe  and  by  Professor  H.  J.  Ryan  in  this 
country;  on  account  of  its  expense,  it  is  used  in  rare  cases 
only. 

When  a  compensating  winding  is  used  in  addition  to  the  inter- 
poles, the  number  of  ampere-turns  on  the  interpoles  is  consider- 
ably reduced,  because  the  compensating  winding  can  be  made  to 
neutralize  the  larger  portion  or  all  of  the  armature  reaction.  In 
such  a  machine  a  much  higher  specific  loading  can  be  allowed  than 
in  an  ordinary  machine  of  the  same  dimensions.  Therefore,  such 
compensated  machines  are  particularly  well  adapted  for  rapidly 
fluctuating  loads,  and  for  sudden  overloads  or  reversals  of  rotation 
in  the  case  of  a  motor. 

Prob.  13.  From  the  following  data  determine  the  ampere-turns 
required  on  each  commutating  pole  of  a  turbo-generator:  The  com- 
mutating poles  are  made  of  cast  steel;  the  average  flux  density  on 
the  face  of  the  interpole  is  6000  maxwells  per  sq.cm.;  the  pole-face 
area  250  sq.cm.;  the  pole  cross-section  160  sq.cm.;  the  radial  length 
of  the  interpole  27  cm.;  the  leakage  factor,  1.5.  The  air-gap  reluctance 
is  2.7  millirels,  the  true  tooth  density  20  kilolines  per  sq.cm.,  the  height 
of  the  tooth  4.5  cm.,  and  the  armature  ampere-turns  per  pole  9500. 

Ans.    About  15,300. 

Prob.  14.    The  rated  current  of  the  machine  in  the  preceding  problem 


CHAP.  IX]     ARMATURE  REACTION  IN  D.C.  MACHINES  175 

is  1200  amp.,  and  in  addition  to  the  interpoles  the  machine  is  to  be 
provided  with  a  compensating  winding.  Show  that  each  interpole 
should  have  at  least  8  turns,  and  each  main  pole  be  provided  with  10 
bars  for  the  compensating  winding,  in  order  to  neutralize  the  armature 
distortion  under  the  main  poles  and  provide  the  proper  commutating 
field. 

Prob.  15.  Machines  provided  with  interpoles  are  very  sensitive 
as  to  their  brush  position.  By  shifting  the  brushes  even  slightly  from 
the  geometrical  neutral  the  terminal  voltage  of  such  a  generator  can  be 
varied  to  a  considerable  extent.  In  the  case  of  a  motor  the  speed  can 
be  regulated  by  this  method,  without  adjusting  the  field  rheostat.  Give 
an  explanation  of  this  "  compounding"  effect  of  the  brush  shift. 

55.  Armature  Reaction  in  a  Rotary  Converter.  The  actual 
currents  of  irregular  form  which  flow  in  the  armature  winding  of  a 
rotary  converter  may  be  considered  as  the  resultants  of  the  direct 
current  taken  from  the  machine  and  of  the  alternating  currents 
taken  in  by  the  machine.  The  resultant  magnetizing  action  upon 
the  field  is  the  same  as  if  these  two  kinds  of  currents  were  flowing 
through  two  separate  windings.  Therefore,  the  armature  reaction 
in  a  rotary  converter  can  be  calculated  by  properly  combining  the 
armature  reactions  of  a  synchronous  motor  and  of  a  direct-current 
generator. 

The  alternating-current  input  into  a  rotary  converter  may  be 
either  at  a  power  factor  of  unity,  if  the  field  excitation  is  properly 
adjusted,  or  the  input  may  have  a  lagging  or  a  leading  component, 
the  same  as  in  the  case  of  a  synchronous  motor.  The  armature 
reaction  due  to  the  energy  component  of  the  input  consists  chiefly 
in  the  distortion  of  the  field,  against  the  direction  of  rotation  of  the 
armature.  But  the  action  of  the  direct  current  is  to  distort  the 
field  in  the  direction  of  rotation,  and  since  the  two  m.m.fs.  are  not 
much  different  from  one  another,  the  resultant  transverse  arma- 
ture reaction  is  very  small.  The  direct  reaction  of  the  direct  cur- 
rent depends  upon  the  position  of  the  brushes,  and  the  direct 
reaction  due  to  the  alternating  currents  is  determined  by  the 
reactive  component  of  the  input,  which  component  may  vary 
within  wide  limits.  Thus,  the  resultant  direct  reaction  of  a  rotary 
converter  may  be  adjusted  to  almost  any  desired  value. 

The  ohmic  drop  in  the  armature  of  a  rotary  converter  has  a 
different  expression  than  in  either  a  direct-current  or  a  synchro- 
nous machine,  because  the  i2r  loss  must  be  calculated  for  the  actual 
shape  of  the  superimposed  currents.  Rotary  converters  are  some- 


176  THE  MAGNETIC  CIRCUIT  [ART.  55 

times  provided  with  interpoles,  in  order  to  improve  the  commu- 
tation and  to  use  a  higher  specific  electric  loading.1 

1  For  further  details  in  regard  to  the  armature  reaction  in  a  rotary  con- 
verter see  Arnold,  Wechselstromtechnik,  Vol.  4  (1904),  Chap.  28;  Pichel- 
mayer,  Dynamobau  (1908),  p.  276;  Standard  Handbook,  index  under  "  Con- 
verter, synchronous,  effective  armature  resistance";  Parshall  &  Hobart, 
Electric  Machine  Design  (1906)  p.  377;  Lamme  and  Newbury,  Interpoles  in 
Synchronous  Converters,  Trans.  Amer.  Inst.  Electr.  Engrs.,  Vol.  29  (1910), 
p.  1625. 


CHAPTER  X 
ELECTROMAGNETIC   ENERGY   AND   INDUCTANCE 

56.  The  Energy  Stored  in  an  Electromagnetic  Field.  Experi- 
ment shows  that  no  supply  of  energy  is  required  to  maintain  a 
constant  magnetic  field.  The  power  input  into  the  exciting  coil  or 
coils  is  in  this  case  exactly  equal  to  that  converted  into  the  i2r  heat 
in  the  conductors,  and  this  power  is  the  same  whether  a  magnetic 
field  is  produced  or  not.  Another  familiar  example  is  that  of  a 
permanent  magnet,  which  maintains  a  magnetic  field  without  any 
supply  of  energy  from  the  outside,  and  apparently  without  any 
decrease  in  its  internal  energy.  Nevertheless,  every  magnetic  field 
has  a  certain  amount  of  energy  stored  within  it,  though  in  a  form 
yet  unknown.  This  is  proved  by  the  fact  that  an  expenditure  of 
energy  is  required  to  increase  the  field,  and,  on  the  other  hand, 
when  the  flux  is  reduced,  some  energy  is  returned  into  the 
exciting  electric  circuit. 

The  conversion  of  electric  into  magnetic  energy  and  vice 
versa  is  accomplished  through  the  e.m.f.  induced  by  the  vary- 
ing flux.  Let  a  magnetic  flux  be  excited  by  a  coil  CC  (Fig.  45) 
supplied  with  current  from  a  source  of  constant  voltage  E,  and  let 
there  be  a  rheostat  r  in  series  with  the  coil.  Let  part  of  the  resist- 
ance in  the  rheostat  be  suddenly  cut  out  in  order  to  increase  the 
current  in  the  coil.  It  will  be  found  that  the  current  rises  to  its 
final  value  not  instantly ;  namely,  when  the  current  increases,  the 
flux  also  increases,  and  in  so  doing  it  induces  in  the  electric  circuit 
an  e.m.f.,  say  e,  which  tends  to  oppose  the  current.  This  et 
together  with  the  iR  drop,  balances  the  voltage  E,  so  that  for  a 
time  the  power  Ei  supplied  by  the  source  is  larger  than  the  power 
i2R  lost  in  the  total  resistance  of  the  circuit.  The  difference, 
Ei  —i2R,  is  stored  in  the  magnetic  field  created  by  the  coil  and  by 
the  other  parts  of  the  circuit.  The  energy  eidt  supplied  by  the 
electric  circuit  during  the  element  of  time  dt  is  converted  into  the 
magnetic  energy  of  the  field,  by  the  law  of  the  conservation  of 

177 


178 


THE  MAGNETIC  CIRCUIT 


[ART.  56 


energy,  while  the  amount  (E —e)idt=i2Rdt  is  converted  into 
heat. 

If  now  the  same  resistance  is  suddenly  introduced  into 
the  circuit,  the  current  gradually  returns  to  its  former  value,  and 
during  this  variable  period  the  i2R  loss  is  larger  than  the  power  Ei 
supplied  by  the  source.  The  applied  voltage  is  in  this  case  assisted 
by  the  voltage  e  induced  by  the  decreasing  field,  the  e.m.f .  e  sup- 
plying part  of  the  i2R  loss. 

To  make  the  matter  more  concrete  let  the  source  of  electric 
energy  be  a  constant-voltage,  direct-current  generator,  driven  by 


FIG.  45. — The    magnetic   field  produced  by  a  coil,  showing  complete  and 

partial    linkages. 

a  steam  turbine.  Let  the  load  consist  of  coils  of  variable  resist- 
ance R,  and  let  the  coils  produce  a  considerable  magnetic  field. 
As  long  as  the  load  current  is  constant,  the  rate  of  the  steam  con- 
sumption is  determined  by  the  i2R  loss  in  the  circuit.  When  the 
current  increases,  the  steam  is  consumed  at  each  instant  at  a 
higher  rate  than  it  would  be  with  a  constant  current  of  the  same 
instantaneous  value.  The  energy  of  steam  is  thus  partly  stored 
in  the  magnetic  field  of  the  coils.  When  the  current  is  returned 
to  its  former  value,  the  steam  consumption  during  the  transitional 


CHAP.  X]  ENERGY  AND  INDUCTANCE  179 

period  is  less  than  that  which  corresponds  to  the  i2R  loss  in  the 
circuit,  so  that  practically  the  same  amount  of  steam  is  saved 
which  was  expended  before  in  increasing  the  magnetic  field. 

These  phenomena  may  be  explained,  or  at  least  expressed  in 
more  familiar  terms,  by  assuming  the  magnetic  field  to  be  due  to 
some  kind  of  motion  in  a  medium  possessed  of  inertia  (Art.  3). 
When  the  field  strength  is  increased  it  becomes  necessary  to  accel- 
erate the  parts  in  motion,  overcoming  their  inertia.  When  the 
field  is  reduced,  the  kinetic  energy  of  motion  is  returned  to  the 
electric  circuit.  One  can  also  conceive  of  the  energy  of  the  mag- 
netic field  to  be  static  and  in  the  form  of  some  elastic  stress. 
Under  this  hypothesis,  when  a  current  increases,  the  magnetic 
stress  also  increases  at  the  expense  of  the  electric  energy.  In 
either  case,  when  the  current  is  constant  no  energy  is  required 
to  maintain  the  field,  any  more  than  to  maintain  a  constant  rota- 
tion in  a  fly-wheel  or  a  constant  stress  in  an  elastic  body. 

It  seems  the  more  probable  that  the  magnetic  energy  of  a 
circuit  is  stored  in  some  kinetic  form,  because  the  current  which 
accompanies  the  flux  is  itself  a  kinetic  phenomenon.  On  the  other 
hand,  it  appears  more  likely  that  electrostatic  energy  is  due  to  some 
elastic  stresses  and  displacements  in  the  medium,  and  thus  it 
may  be  said  to  be  potential  energy.  Electric  oscillations  and 
waves  consist, then, in  periodic  transformations  of  electrostatic  into 
electromagnetic  energy,  or  potential  into  kinetic  energy,  and  vice 
versa,  similar  to  the  mechanical  oscillations  and  waves  in  an 
elastic  body.  In  the  familiar  case  of  current  or  voltage  resonance 
the  total  energy  of  the  circuit  at  a  certain  instant  is  stored  in  the 
form  of  electrostatic  energy  in  the  condenser  (permittor)  con- 
nected into  the  circuit,  or  in  the  natural  permittance  of  the  circuit  ; 
the  current  and  the  magnetic  energy  at  this  instant  are  equal 
to  zero.  At  another  instant,  when  the  current  and  the  magnetic 
field  are  at  their  maximum,  the  energy  stored  is  all  in  the  form 
of  magnetic  energy,  and  the  voltage  across  the  condenser  and 
the  stress  in  the  dielectric  are  equal  to  zero.  An  oscillating 
pendulum  offers  a  close  analogy  to  such  a  system.  The  resistance 
of  the  electric  circuit,  and  the  magnetic  and  dielectric  hysteresis, 
are  analogous  to  the  friction  and  windage  which  accompanies 
the  motion  of  the  pendulum. 

As  it  is  of  importance  hi  mechanical  machine  design  to  know 
the  inertia  of  the  moving  parts  of  a  machine,  so  it  is  often  necessary 


180  THE  MAGNETIC  CIRCUIT  [ART.  57 

in  the  design  and  operation  of  electrical  apparatus  and  circuits  to 
know  the  amount  of  energy  stored  in  the  magnetic  field,  or  the 
electro-magnetic  inertia.  This  inertia  modifies  the  current  and 
voltage  relations  in  the  electric  circuit  in  somewhat  the  same  way 
in  which  the  inertia  of  the  reciprocating  parts  in  an  engine  modi- 
fies the  useful  effort.  In  mechanical  design  a  revolving  part  is 
characterized  by  its  moment  of  inertia  -from  which  the  stored 
energy  can  be  calculated  for  any  desired  speed.  So  in  electrical 
engineering  a  circuit  or  an  apparatus  is  characterized  by  its  electro- 
magnetic inertia  or  inductance,  from  which  it  is  possible  to  calculate 
the  magnetic  energy  stored  in  it  at  any  desired  value  of  the  cur- 
rent. In  this  and  in  the  two  next  chapters  expressions  are  deduced 
for  the  inductance  of  some  of  the  principal  types  of  apparatus 
used  in  electrical  engineering. 

Prob.  1.  A  stationary  electromagnet  attracts  and  lifts  its  armature 
with  a  weight  attached  to  it.  Explain  how  the  energy  necessary  for 
the  lifting  of  the  weight  is  supplied  by  the  electric  circuit. 

Prob.  2.  A  direct-current  generator  driven  by  a  water-wheel  is 
subjected  to  very  large  and  sudden  fluctuations  of  the  load,  which 
the  governor  and  the  gate  mechanism  are  unable  to  follow  properly. 
A  heavy  fly-wheel  on  the  generator  shaft  would  improve  the  operating 
conditions.  Would  a  reactance  coil  in  series  with  the  main  circuit 
achieve  the  same  result,  provided  that  it  could  be  made  large  enough 
to  store  the  required  excess  of  energy  ? 

Prob.  3.  What  experimental  evidence  could  be  offered  to  support 
the  contention  that  the  energy  of  an  electric  circuit  is  contained  in  the 
magnetic  field  linked  with  the  current,  and  not  in  the  current  itself  ? 
The  flow  of  current  is  usually  compared  to  that  of  water  in  a  pipe; 
is  not  all  the  kinetic  energy  stored  in  the  moving  water  itself  and  not 
in  the  surrounding  medium,  and  if  so,  is  a  current  of  electricity  really 
like  a  current  of  water  ? 

Prob.  4.  Describe  in  detail  current  and  voltage  resonance  1  and 
free  electrical  oscillations,  from  the  point  of  view  of  the  periodic  conversion 
of  electromagnetic  into  electrostatic  energy  and  vice  versa,  taking  account 
of  the  dissipation  of  energy  in  the  resistance  of  the  circuit. 

57.  Electromagnetic  Energy  Expressed  through  the  Linkages 
of  Current  and  Flux.  In  order  to  obtain  a  general  expression  for 
the  energy  stored  in  the  magnetic  field  of  an  electric  circuit,  con- 
sider first  a  single  loop  of  wireaa  (Fig.  11)  through  which  a  steady 
electric  current  i  is  flowing.  Let  the  cross-section  of  the  wire  be 
small  as  compared  to  the  dimensions  of  the  loop,  so  that  the  flux 

1  See  the  author's  Experimental  Electrical  Engineering,  Vol.  2,  pp.  17  to  25. 


CHAP.  X]  ENERGY  AND  INDUCTANCE  181 

inside  the  wire  may  be  disregarded.  The  electromagnetic  energy 
possessed  by  the  loop  is  equal  to  the  electrical  energy  spent  in 
establishing  the  current  i  in  the  loop  against  the  induced  e.m.f. 
Let  it  and  $t  be  the  instantaneous  values  of  the  current  in 
amperes  and  the  flux  in  webers  at  a  moment  t  during  the  period  of 
building  up  the  flux,  and  let  et  be  the  instantaneous  applied 
voltage.  Let  the  flux  increase  by  dOt  during  an  infinitesimal 
interval  dt;  then  the  electrical  energy  (in  joules)  supplied  from 
the  source  of  power  to  the  magnetic  field  is 

dW  =itetdt  =it(d$t/dt)dt  =itd®t, 

where—  et=  —dtf>t/dt  is  the  instantaneous  e.m.f.  induced  in  the 
loop  by  the  changing  flux.  The  total  energy  supplied  from  the 
electrical  source  during  the  period  of  building  up  the  field  to  its 
final  value  4>  is 


TF=  I   itd$t (98) 

In  a  medium  of  constant  permeability  the  integration  can  be 
easily  performed,  because  the  flux  is  proportional  to  the  current, 
or,  according  to  eq.  (2)  in  Art.  5,  $t  =(Pit,  where  (P  is  the  permeance 
of  the  magnetic  circuit,  in  henrys.  Eliminating  by  means  of  this 
relation  either  it  or  <Dt  from  eq.  (98)  we  can  obtain  any  one  of  the 
following  three  expressions  for  the  electromagnetic  energy  stored 
in  the  loop : 


(99) 


In  the  first  form,  eq.  (99)  expresses  the  fact  that  the  magnetic 
energy  stored  in  a  loop  is  equal  to  one-half  the  product  of  the  cur- 
rent by  the  flux;  in  the  second  form,  it  shows  that  the  stored 
energy  is  proportional  to  the  square  of  the  current  and  to  the  per- 
meance of  the  magnetic  circuit.  Both  forms  are  of  importance  in 
practical  applications. 

Take  now  the  more  general  case  of  a  coil  of  n  turns  (Fig.  45)  ; 
the  flux  which  links  with  a  part  of  the  turns  is  now  of  a  magnitude 
comparable  with  that  of  the  flux  which  links  with  all  the  turns  of 
the  coil.  We  shall  consider  the  complete  linkages  and  the  partial 
linkages  separately.  Consider  first  the  energy  due  to  the  flux 


182  THE   MAGNETIC   CIRCUIT  [ART.  57 

which  links  with  all  the  turns  of  the  coil.      The  e.m.f.  induced  in 
the  coil  by  this  flux,  when  the  current  changes,  is  —  et  =  —n(d0t/df), 
and  the  relation  between  the  current  and  the  flux  is  0t=(Pcnit} 
where  (Pc  is  the  permeance  of  the  path  of  the  complete  linkages. 
By  repeating  the  reasoning  given  above  in  the  case  of  a  single  loop 
we  find  that 

Wc=%ni#c,      .....     .     .     (100) 

or 

......     (lOOa) 


where  the  subcript  c  signifies  that  the  quantities  refer  to  the  com- 
plete linkages  only  (Fig.  45)  .  Two  forms  only  are  retained,  being 
those  that  are  of  the  most  practical  importance. 

The  energy  of  the  partial  linkages  is  calculated  in  a  similar 
manner.  Let  A$P  be  a  small  annular  flux  (Fig.  45)  which  links 
with  Up  turns  of  the  coil,  where  np  may  be  an  integer  or  a  fraction. 
For  these  turns  the  linkage  with  A$P  is  a  complete  linkage,  while 
for  the  external  (n—np)  turns  it  is  no  linkage  at  all  and  represents 
no  energy,  because  no  e.m.f.  is  induced  by  A®?  in  the  turns  external 
to  it.  Thus,  the  energy  due  to  the  flux  A$P,  according  to  eqs. 
(100)  and  (lOOa),  is  equal  to  \nviA^P,  or  to  %nP2i24(PP.  The 
total  energy  of  the  partial  linkages  is  the  sum  of  such  expressions, 
over  the  whole  flux  $P,  or 

WP=%iI>npA$p,     ........     (101) 

or 

WP=%i22nP24(PP.     .     .     .....     (lOla) 

The  total  energy  of  the  coil  is 

A$Pl      .....     (102) 


or 

W=^i2[n2(Pc  +  I,nP2A(PPl      ....     (102a) 

where  the  first  term  on  the  right-hand  side  refers  to  the  complete 
linkages  and  the  second  to  the  partial  linkages  of  the  flux  and  the 
current.  In  these  expressions  the  current  is  in  amperes,  the  fluxes 
in  webers,  the  permeances  in  henrys,  and  the  energy  in  joules 
(watt-seconds)  .  If  other  units  are  used  the  corresponding  numeri- 
cal conversion  factors  must  be  introduced. 


CHAP.  X]  ENERGY  AND  INDUCTANCE  183 

Some  new  light  is  thrown  upon  these  relations  by  using  the 
m.m.f  .  M  instead  of  the  ampere-turns  ni.  Namely,  eqs.  (102)  and 
(102a)  become: 

1fHEtfA*-2iffJ*4      ....    (103) 

or 

.       .     .     .     (103a) 


These  expressions  are  analogous  to  those  for  the  energy  stored  in 
an  electrostatic  circuit,  viz.,  %EQ,  and  %E2C  (see  The  Electric  Cir- 
cuit). The  m.m.f.  Mc  is  analogous  to  the  e.m.f.  E;  the  magnetic 
flux  0C  is  analogous  to  the  electrostatic  flux  Q,  and  the  permeance 
(Pc  is  analogous  to  the  permittance  C. 

We  can  assume  as  a  fundamental  law  of  nature  the  fact  that 
with  a  given  steady  current  the  magnetic  field  is  distributed  in  such 
a  way  that  the  total  electromagnetic  energy  of  the  system  is  a 
maximum.  All  known  fields  obey  this  law,  and,  in  addition,  it  can 
be  proved  by  the  higher  mathematics.  Eq.  (102a)  shows  that  this 
law  is  fulfilled  when  the  sum  nc2(Pc  +  2nP2(Ppisa,  maximum.  When 
the  partial  linkages  are  comparatively  small,  the  energy  stored  is  a 
maximum  when  the  permeance  (Pc  of  the  paths  of  the  total  linkages 
is  a  maximum.  This  fact  is  made  use  of  in  the  graphical  method 
of  mapping  out  a  magnetic  field,  in  Art.  41  above. 

Prob.  5.  The  no-load  saturation  curve  of  an  8-pole  electric  gen- 
erator is  a  straight  line  such  that  when  the  useful  flux  is  10  megalines 
per  pole  the  excitation  is  7200  amp  .-turns  per  pole;  the  leakage  factor 
is  1.2.  Show  that  at  this  excitation  there  is  enough  energy  stored 
in  the  field  to  supply  a  small  incandescent  lamp  with  power  for  a  few 
minutes. 

Prob.  6.  Explain  the  function  and  the  diagram  of  connections  of 
a  field-discharge  switch. 

Prob.  7.  Prove  that  the  magnetic  energy  stored  in  an  apparatus 
containing  iron  is  proportional  to  the  area  between  the  saturation  curve 
and  the  axis  of  ordinates.  The  saturation  curve  is  understood  to  give 
the  total  flux  plotted  against  the  exciting  ampere-turns  as  abscissae. 
Hint:  See  Art.  16. 

Prob.  8.  Deduce  expression  (102a)  directly,  by  writing  down  an 
expression  for  the  total  instantaneous  e.m.f.  induced  in  a  coil  (Fig.  45). 

Prob.  9.  Explain  the  reason  for  which,  in  the  formulae  deduced  above, 
it  is  permissible  to  consider  n  to  be  a  fractional  number. 

58.  Inductance  as  the  Coefficient  of  Stored  Energy,  or  the 
Electrical  Inertia  of  a  Circuit.  Eq.  (102a)  shows  that  in  a  mag- 
netic circuit  of  constant  permeability  the  stored  energy  is  proper- 


184  THE  MAGNETIC  CIRCUIT  [ART.  58 

tional  to  the  square  of  the  current  which  excites  the  field.  The 
coefficient  of  proportionality,  which  depends  only  upon  the  form 
of  the  circuit  and  the  position  of  the  exciting  m.m.f.,  is  defined 
as  the  inductance  of  the  electric  circuit.  The  older  name  for 
inductance  (is  the  coefficient  of  self-induction.  It  is  assumed 
here  that  the  magnetic  circuit  is  excited  by  only  one  electric  cir- 
cuit, so  that  there  is  no  mutual  inductance.  Thus,  by  definition 


....     (104) 
where  the  inductance  is 


.     .    .    .     .     (105) 
or,  replacing  the  summation  by  an  integration, 

L  =n*(Pc  +  CnnP2d(PP.  (106) 

JQ 

Since  the  permeances  in  eq.  (102a)  are  expressed  in  henrys,  and 
the  numbers  of  turns  are  numerics,  the  inductance  L  in  the  defin- 
ing eqs.  (105),  or  (106),  is  also  in  henrys.  If  the  permeances  are 
measured  in  millihenrys  or  in  perms,  the  inductance  L  is  measured 
in  the  same  units.  As  a  matter  of  fact,  the  henry  was  originally 
adopted  as  a  unit  of  inductance,  and  only  later  on  was  applied  to 
permeance.1 

In  some  cases  it  is  convenient  to  replace  the  actual  coil  (Fig.  45) 
by  a  fictitious  coil  of  an  equal  inductance,  and  of  the  same  number  of 
turns,  but  without  partial  linkages.  Let  (Peq  be  the  permeance  of 
the  complete  linkages  of  this  fictitious  coil;  then,  by  definition, 
eqs.  (105)  and  (106)  become 

L=n2(Peq.         ......     (106a) 

This  expression  is  used  when  the  permeance  of  the  paths  is  calcu- 
lated from  the  results  of  experimental  measurements  of  inductance, 
because  in  this  case  it  is  not  possible  to  separate  the  partial 
linkages.  Use  is  made  of  formula  (106a)  in  chapter  XII,  in  cal- 
culating the  inductance  of  armature  windings. 

1  The  use  of  the  henry  as  a  unit  of  permeance  was  proposed  by  Professor 
Giorgi.  See  Trans.  Intern.  Elec.  Congress  at  St.  Louis  (1904),  Vol.  1,  p.  136. 
The  connection  between  inductance  and  permeance  seems  to  have  been  first 
established  by  Oliver  Heaviside;  see  his  Electromagnetic  Theory  (1894),  Vol. 
1,  p.  31. 


CHAP.  X]  ENERGY  AND  INDUCTANCE  185 

The  inductance  L  is  related  in  a  simple  manner  to  the  electro- 
motive force  induced  in  the  exciting  electric  circuit  when  the  cur- 
rent varies  in  it.  Namely,  the  electric  power  supplied  to  the  cir- 
cuit or  returned  from  the  circuit  to  the  source  is  equal  to  the  rate 
of  change  of  the  stored  energy,  so  that  we  have  from  eq.  (104) 

dW/dt=i(-e)=Li(di/dt), 
or,  canceling  i, 

e  =  -L(di/d£),       . .    ,'  .     (107) 

The  sign  minus  is  used  because  e  is  understood  to  be  the  induced 
e.m.f.  and  not  that  applied  at  the  terminals  of  the  circuit.  There- 
fore, when  dW/dt  is  positive,  that  is,  when  the  stored  energy 
increases  with  the  time,  e  is  induced  in  the  direction  opposite  to  that 
of  the  flow  of  the  current,  and  hence  by  convention  is  considered 
negative.  Inductance  is  sometimes  defined  by  eq.  (107),  and  then 
eqs.  (104)  and  (105)  are  deduced  from  it.  The  definition  of  L  by 
the  expression  for  the  electromagnetic  energy  seems  to  be  a  more 
logical  one  for  the  purpose  of  this  treatise,  while  the  other  defini- 
tion in  terms  of  the  induced  e.m.f.  is  proper  from  the  point  of 
view  of  the  electric  circuit. 

Looking  upon  the  stored  magnetic  energy  as  due  to  some  kind 
of  a  motion  in  the  medium,  eq.  (104)  suggests  the  familiar  expres- 
sions %mv2  and  %Kaj2  for  the  kinetic  energy  of  a  mechanical  system. 
Taking  the  current  to  be  analogous  to  the  velocity  of  motion,  the 
inductance  becomes  analogous  to  mechanical  mass  and  moment  of 
inertia.  The  larger  the  electromagnetic  inertia  L  the  more  energy 
is  stored  with  the  same  current.  Equation  (107)  also  has  its 
analogue  in  mechanics,  namely  in  the  familiar  expressions  mdv/dt 
and  Kdco/dt  for  the  accelerating  force  and  torque  respectively. 
The  e.m.f.  e  represents  the  reaction  of  the  circuit  upon  the  source 
of  power  when  the  latter  tends  to  increase  i  the  rate  of  flow  of  elec- 
tricity. While  these  analogies  should  not  be  carried  too  far,  they 
are  helpful  in  forming  a  clearer  picture  of  the  electromagnetic 
phenomena. 

The  role  of  inductance,  L,  in  the  current  and  voltage  relations  of 
alternating-current  circuits  is  treated  in  detail  in  the  author's 
Electric  Circuit.  In  this  book  inductance  is  considered  from  the 
point  of  view  of  the  magnetic  circuit,  i.e.,  as  expressed  by  eqs.  (104) 
to  (106) .  In  the  next  two  chapters  the  values  of  inductance  are 


186  THE  MAGNETIC  CIRCUIT  [ART.  58 


calculated  for  some  important  practical  cases,  from  the  forms  and 
the  dimensions  of  the  magnetic  circuits,  using  the  fundamental 
equations  (104),  (105)  and  (106).  The  reader  will  see  that  the 
problem  is  reduced  to  the  determination  of  various  permeances  and 
fluxes;  hence,  it  presents  the  same  difficulties  with  which  he  is 
already  familiar  from  the  study  of  Chapters  V  and  VI. 

Inductance  of  electric  circuits  in  the  presence  of  iron.  When 
iron  is  present  in  the  magnetic  circuit,  three  cases  may  be  distin- 
guished : 

(1)  The  reluctance  of  the  iron  parts  is  negligible  as  compared 
to  that  of  the  rest  of  the  circuit ; 

(2)  The  reluctance  of  the  iron  parts  is  constant  within  the 
range  of  the  flux  densities  used ; 

(3)  The  reluctance  of  the  iron  parts  is  considerable,  and  is 
variable. 

In  the  first  two  cases,  eqs.  (104),  (105)  and  (106)  hold  true,  and 
the  inductance  can  be  calculated  from  the  constant  permeances  of 
the  magnetic  circuit.  In  the  third  case,  inductance,  if  used  at  all, 
must  be  separately  defined,  because  eq.  (1026)  does  not  hold  when 
the  permeance  of  the  circuit  varies  with  the  current.  The  equa- 
tion of  energy  is  in  this  case 

W=n  (\d0,c  +  2  rnpitd(4$tp).  (108) 

Jo  Jo 

This  equation  is  deduced  by  the  same  reasoning  as  eq.  (98) . 

The  following  three  definitions  of  inductance  are  used  by  differ- 
ent authors  when  the  reluctance  of  a  magnetic  circuit  is  variable : 
(a)  the  expressions  (104)  and  (108)  are  equated  to  each  other,  and 
L  is  calculated  separately  for  each  final  value  i  of  the  current. 
Thus  L  is  variable,  and  neither  eq.  (105)  nor  (107)  hold  true,  (b) 
L  is  defined  from  eq.  (107) ;  in  this  case  neither  eq.  (104)  nor  (105) 
are  fulfilled,  (c)  L  is  defined  at  a  given  current  by  eq.  (105)  so 
that  Li  represents  the  sum  of  the  linkages  of  the  flux  and  the  cur- 
rent. Therefore  eq.  (107)  becomes  e  =  —d(Li)/dt,  and  dW  =id(Li) . 
With  each  of  the  three  definitions  L  is  variable,  and  therefore  is  not 
very  useful  in  applications.  The  author's  opinion  is  that  when  the 
permeance  of  the  circuit  is  variable,  L  should  not  be  introduced  at 
all,  but  the  original  equation  of  energy  (108)  be  used  directly.  Or 
else  in  approximate  calculations,  a  constant  value  of  L  can  be 
used,  calculated  for  some  average  value  of  i  or  0. 


CHAP.  X]  ENERGY  AND  INDUCTANCE  187 

Prob.  10.  It  is  desired  to  make  a  standard  of  inductance  of  one 
millihenry  by  winding  uniformly  one  layer  of  thin  flat  conductor  upon 
a  toroidal  wooden  ring  of  circular  cross-section.  How  many  turns  are 
needed  if  the  diameter  of  the  cross-section  of  the  ring  is  10  cm.  and 
the  mean  diameter  of  the  ring  itself  is  50  cm.?  Ans.  400. 

Prob.  11.  An  iron  ring  of  circular  cross-section  is  uniformly  wound 
with  n  turns  of  wire,  the  total  thickness  of  the  winding  being  t;  the 
mean  diameter  of  the  ring  is  D  and  the  radius  of  its  cross-section  r. 
What  is  the  inductance  of  the  apparatus,  assuming  the  permeability 
of  the  iron  to  be  constant  and  equal  to  1500  times  that  of  the  air? 
Hint:  d6>p  =  tt2n(r+x)dx/nD',  np=n(x/t). 

Ans.     1.25(n2/jD)  [1500r2  +  t($r  +  #)]  X  10~8  henry. 

Prob.  12.     A  ring  is  made  of  non-magnetic    material,  and  has  a 

rectangular  cross-section  of  dimensions  b  and  h]    the  mean  diameter 

of  the  ring  is  D.    It  is  uniformly  wound  with  n  turns  of  wire,  the  total 

thickness  of  the  winding  being  t.     What  is  the  inductance  of  the  winding? 

Ans.     0.4(n2/#)  \bh  +  $t(b  +h +3f)]lQ-*  millihenry. 

Prob.  13.  The  ring  in  the  preceding  problem  has  the  following 
dimensions:  D  =  50  cm.;  /i  =  b  =  10  cm.;  it  is  to  be  wound  with  a  con- 
ductor 3  mm.  thick  (insulated).  How  many  turns  are  required  in  order 
to  get  an  inductance  of  0.43  of  a  henry?  Hint :  Solve  by  trials,  assum- 
ing reasonable  values  for  t;  the  number  of  turns  per  layer  decreases  as 
the  thickness  of  the  winding  increases.  Ans.  About  5300. 

Prob.  14.  It  is  desired  to  design  a  choke  coil  which  will  cause  a 
reactive  drop  of  250  volts,  at  10  amp.  and  50  cycles.  The  cross-section 
of  the  core  (Fig.  12)  is  120  sq.cm.,  and  the  mean  length  of  the  path 
130  cm.;  the  maximum  flux  density  in  the  iron  must  be  not  over  7 
kl.  per  sq.  cm.  What  is  the  required  number  of  turns  and  the  length 
of  the  air-gap  in  the  core?1  Ans.  150;  3.7  mm. 

Prob.  15.  An  electrical  circuit,  which  consists  of  a  Leyden  jar 
battery  of  0.01  mf.  capacity  and  of  a  coil  having  an  inductance  of  10 
millihenrys,  undergoes  free  electrical  oscillations  in  such  a  way  that  the 
maximum  instantaneous  voltage  across  the  condenser  is  10,000  volts. 
What  is  the  current  through  the  inductance  one-quarter  of  a  cycle 
later,  neglecting  any  loss  of  energy  during  the  interval  ? 

Ans.     10  amp. 

Prob.  16.  Suggest  a  practical  experiment  which  would  prove  directly 
that  the  stored  electromagnetic  energy  is  proportional  to  the  square 
of  the  current. 

Prob.  17.  Prove  that  the  inductance  of  a  coil  of  given  external 
dimensions  is  proportional  to  the  square  of  the  number  of  turns,  taking 
into  account  the  complete  and  the  partial  linkages.  Show  that  the 
ohmic  resistance  of  the  coil  is  also  proportional  to  the  square  of  the 
number  of  turns,  provided  that  the  space  factor  is  constant. 

NOTE  1.    The   theoretical   calculation   of  the  ~  inductance   of   short 

1  For  a  complete  design  of  a  reactive  coil  see  G.  Kapp,  Transformers  (1908), 
p.  105. 


188  THE   MAGNETIC   CIRCUIT  [ART.  58 

straight  coils  and  loops  of  wire  in  the  air  is  rather  complicated,  because 
of  the  mathematical  difficulties  in  expressing  the  permeances  of  the 
paths.  Those  interested  in  the  subject  will  find  ample  information 
in  Rosa  and  Cohen's  Formula;  and  Tables  for  the  Calculation  of  Mutual 
and  Self-Inductance,  in  the  Bulletins  of  the  Bureau  of  Standards,  Vol. 
5  (1908),  No.  1.  The  article  contains  also  quite  a- complete  bibliography 
on  the  subject.  See  also  Orlich,  Kapazit'dt  und  Inductivit'dt  (1909), 
p.  74  et  seq. 

Note  2.  In  the  formulae  deduced  in  this  and  in  the  two  following 
chapters,  it  is  presupposed  that  the  current  is  distributed  uniformly 
over  the  cross-section  of  the  conductors.  Such  is  the  case  in  conductors 
of  moderate  size  and  at  ordinary  commercial  frequencies,  unless  perchance 
the  material  is  itself  a  magnetic  substance.  With  high  frequencies, 
or  with  conductors  of  unusually  large  transverse  dimensions,  as  also  with 
conductors  of  a  magnetic  material,  the  current  is  not  distributed  uni- 
formly over  the  cross-section  of  the  conductors,  the  current  density 
being  higher  near  the  periphery.  The  result  is  that,  as  the  frequency 
increases,  the  inductance  becomes  lower  and  the  ohmic  resistance  higher. 
This  is  known  as  the  skin  effect.  For  an  explanation,  for  a  mathematical 
treatment  in  a  simple  case,  and  for  references  see  Heinke,  Handbuch 
der  Elektrotechnik,  Vol.  1  (1904)  part  2,  pp.  120  to  129.  Tables  and  for- 
mulas will  be  found  in  the  Standard  Handbook,  and  in  Foster's  Pocket- 
Book.  See  also  C.  P.  Steinmetz,  Alternating-Current  Phenomena  (1908), 
pp.  206-208,  and  his  Transient  Electric  Phenomena  (1909),  Section  III, 
Chapter  VII;  Arnold,  Die  Wechselstromtechnik,  Vol.  1  (1910),  p.  564;  A. 
B.  Field,  Eddy  Currents  in  Large  Slot-wound  Conductors,  Trans.  Amer. 
Inst.  Electr.  Engrs.,  Vol.  24  (1905),  p.  761. 


CHAPTER  XI. 


Let  a  direct  current  of  i 


THE   INDUCTANCE    OF   CABLES   AND    OF   TRANS- 
MISSION   LINES. 

59.  The  Inductance  of  a  Single-phase  Concentric  Cable.  Let 
Fig.  46  represent  the  cross-section  of  a  concentric  cable,  which 
consists  of  an  inner  core  A  and  of  an  external  annular  conductor 
D,  with  some  insulation  C  between  them.  Let  the  radii  of  the 
conductors  be  a,  b,  and  c,  respectively.  The  insulation  outside  of 
D  and  the  sheathing  are  not  shown, 
amperes  flow  through  the  inner 
conductor  away  from  the  reader 
and  return  through  the  outer 
conductor.  The  magnetic  field 
produced  by  this  current  links 
with  the  current,  and  for  reasons 
of  symmetry  the  lines  of  force 
are  concentric  circles.  The  field 
is  confined  within  the  cable, 
because  outside  the  external  con- 
ductor D  the  m.m.f.  isi  —  i=0. 

In  the  space  between  the  two 
conductors  the  lines  of  force  are 
linked  with  the  whole  current, 
and  since  there  is  but  one  turn, 
the  m.m.f.  is  equal  to  i.  The  length  of  a  line  of  force  of  a  radius  x 
cm.  is  27rzsothat  the  magnetic  intensity  isH  =i/2nx,  amp.  turns  per 
cm.,  the  corresponding  flux  density  B=/j.i/2nx  maxwells  per  sq.cm. 
Thus,  the  flux  density  decreases  inversely  as  the  distance  from  the 
center;  it  is  represented  by  the  ordinates  of  the  part  qr  of  a  hyper- 
bola. 

In  the  space  within  the  inner  conductor  A,  a  line  of  force  of 
radius  x  is  linked  with  a  current  ix=i(nx2/na?)  =i(x/a)2,  provided 

189 


Fig.  46— The  magnetic  field  within 
a  concentric  cable. 


190  THE   MAGNETIC   CIRCUIT  [ART.  59 

that  the  current  is  distributed  uniformly  over  the  cross-section  of 
the  conductor.  The  length  of  the  line  of  force  is  2xx,  so  that 
H=ix/2nx=xi/(2na2),  and  B=fix  i/(2na2).  Thus,  in  this  part 
of  the  field  the  flux  density  increases  as  the  distance  from  the 
center  and  is  represented  by  the  straight  line  Oq. 

In  the  space  inside  the  conductor  D,  a  line  of  force  of  a 
radius  x  is  linked  with  the  current  —  i(x2  —  62)/(c2—  b2)  of  the 
external  conductor  and  with  the  current  +i  of  the  internal  con- 
ductor, or  altogether  with  the  current  ix=i(c2—x2)/(c2—b2). 
Consequently,  here,  the  flux  density  is  represented  by  the 
hyperbola  rs,  the  equation  of  which  is 

B=fjLix/27tx  =  fjii(c2/x  -x)/[2n(c2  -62)]. 

The  curve  Oqrs  gives  a  clear  physical  picture  of  the  field  dis- 
tribution in  the  cable,  and  helps  one  to  understand  the  linkages 
which  enter  into  the  calculation  of  the  inductance  of  the  cable. 

The  inductance  of  the  cable  is  calculated  according  to  the 
fundamental  formula  (106),  the  complete  linkages  being  in  the 
space  between  the  two  conductors,  and  the  partial  linkages  being 
within  the  space  occupied  by  the  conductors  themselves.  Con- 
sider a  piece  of  the  cable  one  centimeter  long.  The  permeance  of  a 
tube  of  force  of  a  radius  x  and  of  a  thickness  dx  is  fj.dx/2nx,  so  that 
the  permeance  of  the  complete  linkages  is, 

LC'=(PC'=  CbfjLdx/2nx  =  (fjL/2n)Ln(b/a)  perm/cm.,  (109) 


where  Ln  is  the  symbol  for  natural  logarithms.  In  this  case  the 
permeance  is  equal  to  the  inductance  because  the  number  of  turns 
n  =  l.  The  sign  "prime"  indicates  that  the  quantities  Lcr  and 
(Pef  refer  to  a  unit  length  of  the  cable. 

For  the  space  inside  the  inner  conductor  np  =  (x/a)2,  this  being 
the  fraction  of  the  current  with  which  the  line  of  force  of  radius  x 
is  linked.  Hence,  the  part  of  the  inductance  of  the  cable  due  to 
the  field  inside  the  conductor  A  is 


LA'  =  (/£/27r)  C*(x/a)*(dx/x)  =n/8n  =0.05  perm/cm.      (110) 
^o 

This  formula  shows  that  the  part  of  the  inductance  due  to  the 
field  within  the  inner  conductor  is  independent  of  the  radius  of  the 


CHAP.  XI]      INDUCTANCE  OF  TRANSMISSION  LINES  191 

conductor,  and  is  always  equal  to  0.05  perm,  per  cm.,  or  0.05  milli- 
henry per  kilometer  length  of  the  cable. 

The  exact  expression  for  the  part  of  the  inductance  of  the  cable 
LD'  due  to  the  linkage  within  the  outer  conductor  is  given  in 
problem  10  below.  The  formula  is  rather  complicated  for  prac- 
tical use,  especially  in  view  of  the  fact  that  this  part  of  the  induct- 
ance is  comparatively  small,  because  the  flux  density  on  the  part 
rs  of  the  curve  is  small.  It  is  more  convenient,  therefore,  to  make 
simplifying  assumptions,  when  the  thickness  t  of  the  outer  con- 
ductor is  small  as  compared  to  b.  Namely,  the  length  of  all  the 
paths  within  the  outer  conductor  may  be  assumed  to  be  equal 
to  2nb,  so  that  the  permeance  of  an  infinitesimal  path  of  a  radius  x 
and  thickness  dx  nearly  equals  fjidx/2xb.  Furthermore,  the  volume 
of  the  current  in  the  outer  conductor,  between  the  radii  b  and  x> 
may  be  assumed  to  be  proportional  to  the  distance  z— &,  and 
hence  equal  to  i (x  —b}/(c  —b).  A  line  of  force  of  a  radius  x  is  linked 
therefore  with  the  whole  current  i  in  the  inner  conductor  and  with 
the  above-stated  part  of  the  current  in  the  outer  conductor,  and, 
since  the  currents  flow  in  opposite  directions,  this  line  of 
force  is  linked  altogether  with  the  current  i(c—x)/(c—b).  Hence, 
it  is  linked  with  np  =  (c  —x)/(c  —b)  turns.  Thus, the  inductance  of 
the  cable,  due  to  the  outer  partial  linkages,  is,  in  the  first  approxi- 
mation, 

LD'=n/(2nbt2)  C\c-x)2dx=^t/b  perm/cm.        .     (Ill) 

"b 

If  a  closer  approximation  is  desired,  it  is  convenient  to  expand 
eq.  (114)  in  ascending  powers  of  t/b,  as  is  explained  in  problem  11. 
The  result  is 

LD'  =rV/^[^  ~ro(^/^)2^~  A(V^)3  •  •  •]  perm/cm.   .     (112) 

It  will  thus  be  seen  that  eq.  (Ill)  is  an  accurate  approximation, 
because  eq.  (112)  contains  in  the  parentheses  no  term  with  the 
first  power  of  the  ratio  t/b. 

Thus,  the  total  inductance  of  a  concentric  cable,  I  kilometers 
long,  is 

L  =[0.46  Log10  (b/a)  +0.05+LD']Z  millihenrys,    .     (113) 

where  LD'  is  given  by  eqs.  (Ill),  (112),  or  (114),  according  to  the 
accuracy  desired.  Expressions  (110)  and  (114)  are  correct  only  at 
low  frequencies,  such  as  are  used  for  power  transmission.  With 


192  THE  MAGNETIC  CIRCUIT  [ART.  59 

very  high  frequencies,  the  skin  effect  becomes  noticeable,  that  is, 
the  current  in  the  inner  conductor  is  forced  outward  and  that  in 
the  outer  conductor  inward.  In  the  limit,  when  the  frequency 
is  infinite,  the  currents  are  concentrated  on  the  opposing  surfaces 
of  the  conductors,  and  the  partial  linkages  are  equal  to  zero. 
Thus,  each  of  the  expressions  in  question  must  be  multiplied  by  a 
variable  coefficient  k  which,  for  a  given  cable,  is  a  function  of 
the  frequency.  At  ordinary  frequencies  k  =  1,  and  gradually 
approaches  zero  as  the  frequency  increases  to  infinity.1 

Prob.  1.  A  concentric  cable  is  to  be  designed  for  750  amperes,  the 
current  density  to  be  about  2.2  amp.  per  gross  sq.mm.,  and  the  thickness 
of  the  insulation  between  the  conductors  to  be  6  mm.  What  are  the 
dimensions  of  the  conductors  assuming  them  to  be  solid,  that  is,  not 
stranded?  The  fact  that  they  are  in  reality  stranded  is  taken  care  of 
in  the  permissible  current  density. 

Ans.    a  =  10.5 ;  b  =  16.5 ;  c  =  19.6  mm. 

Prob.  2.  Plot  the  curve  Oqrs  of  distribution  of  the  flux  density 
in  the  cable  given  in  the  preceding  problem. 

Ans.    At  x=  a,  B=  143;  at  x  =  b,  B=  91  maxwells  per  sq.cm. 

Prob.  3.  What  is  the  total  flux  in  megalines  per  kilometer  of  the 
cable  specified  in  the  two  preceding  problems?  Ans.  15.1. 

Prob.  4.  Show  how  to  plot  the  cilrve  of  the  distribution  of  flux  density 
in  a  three-phase  concentric  cable,  at  some  given  instantaneous  values 
of  the  three  currents. 

Prob.  5.  What  is  the  inductance  of  a  25-km.  cable  in  which  the 
diameter  of  the  inner  conductor  is  12  mm.,  the  thickness  of  insulation 
is  3  mm.,  and  the  dimension  c  is  such  that  the  current  density  in  the 
outer  conductor  is  10  per  cent  higher  than  in  the  inner  one  ? 

Ans.     25  [0.0810 +  0.050 +  0.0122]  =  3.58  millihenry. 

Prob.  6.  A  cable  consists  of  three  concentric  cylinders  of  negligible 
thickness;  the  radii  of  the  cylinders  are  rv,  r2,  and  r3,  beginning  with 
the  inner  one.  What  is  the  inductance  in  millihenrys  per  kilometer, 
when  a  current  flows  through  the  inner  cylinder  and  returns  equally 
divided  through  the  two  others?  , 

Ans.    0.46  [log  (r2/rO  +0.25  log  (r3/r2) .] 

Prob.  7.  In  the  cable  given  in  the  preceding  problem  the  total 
current  i  flows  through  the  middle  cylinder,  the  part  mi  returns  through 
the  inner  cylinder,  and  the  rest,  ni,  returns  through  the  outer  one.  What 
is  the  total  inductance  per  kilometer  of  length? 

Ans.      0.46  [m2  log  (r,/rO  +n2  log  (r./r,)]. 

1  For  the  field  distribution  in  and  the  inductance  of  non-concentric  cables 
see  Alex.  Russell,  Alternating  Currents,  Vol.  1  (1904),  Chap.  XV;  for  the 
reactance  of  armored  cables  see  J.  B.  Whitehead,  "  The  Resistance  and  React- 
ance of  Armored  Cables,  Trans.  Amer.  Inst.  Electr.  Engrs.,  Vol.  28  (1909), 
p.  737. 


CHAP.  XI]      INDUCTANCE  OF  TRANSMISSION  LINES  193 

Prob.  8.  At  what  ratio  of  b  to  a  in  Fig.  46  is  the  magnetic  energy 
stored  within  the  inner  conductor  equal  to  that  stored  between  the 
two  conductors  ?•  Ans.  1.28. 

Prob.  9.  It  is  required  to  replace  the  solid  inner  conductor  A  in 
Fig.  46  by  an  infinitely  thin  shell  of  such  a  radius  a'  that  the  total 
inductance  of  the  cable  shall  remain  the  same.  What  is  the  radius  of 
the  shell  ?  Hint  :  (  ft  /2n)  Ln  (a  /a')  =  /I/STT. 

Ans.     a'/a  =  £-°'25=0.779. 

Prob.  10.  Prove  that  the  part  of  the  inductance  due  to  the  linkages 
within  the  outer  conductor  in  Fig.  46  is  expressed  by 


;r_i(3^+^-4c^)].     .     .     (114) 

Hint:  dVp^ndx/lnx;  np  =  l-n(x2-b*)/7t(c*/b2). 

Prob.  11.  Deduce  eq.  (112)  from  formula  (114),  assuming  the 
ratio  t/b  to  be  small  as  compared  to  unity.  Hint:  Put  c  =  b(l+y) 
where  y  =  t/b  is  a  small  fraction.  Expand  Ln(l-y)  into  an  infinite 
series,  and  omit  in  the  numerator  of  eq.  (114)  all  the  terms  above  f/5; 
expand  (c2  —  62)  in  the  denominator  in  ascending  powers  of  y,  and  divide 
the  numerator  by  this  polynomial. 

60.  The  Magnetic  Field  Created  by  a  Loop  of  Two  Parallel 

Wires.  Let  Fig.  47  represent  the  cross-section  of  a  single-phase 
or  direct-current  transmission  line,  the  wires  being  denoted  by 
A  and  B.  With  the  directions  of  the  currents  in  the  wires  shown 
by  the  dot  and  the  cross,  the  magnetic  field  has  the  directions 
shown  by  the  arrow-heads,  one-half  of  the  flux  linking  with  each 
wire.  Before  calculating  the  inductance  of  the  loop  it  is  instruct- 
ive to  get  a  clear  picture,  quantitative  as  well  as  qualitative,  of  the 
field  itself. 

The  field  distribution  is  symmetrical  with  respect  to  the  line 
AB  and  the  axis  00'  .  The  whole  flux  passes  in  the  space  between 
the  wires,  so  as  to  be  linked  with  the  m.m.f  .  which  produces  it,  and 
then  extends  to  infinity  on  all  sides.  The  flux  density  is  at  its 
maximum  near  the  wires  and  gradually  decreases  toward  00' 
and  toward  ±  oo,  as  is  shown  by  the  curve  pqsts'q'p'.  The  ordi- 
nates  of  this  curve  represent  the  flux  densities  at  the  various  points 
of  the  line  passing  through  the  centers  of  the  wires.  The  reason 
for  which  the  flux  density  is  larger  near  the  wires  is  that  the  path 
there  is  shorter,  although  the  m.m.f.  acting  along  all  the  paths 
is  the  same.  This  m.m.f.  is  numerically  equal  to  the  current  i  in 
the  wires,  the  number  of  turns  being  equal  to  one. 

It  is  proved  below  that  the  magnetic  paths  outside  the  conduc- 


194 


THE  MAGNETIC  CIRCUIT 


[ART.  60 


tors  themselves  are  eccentric  circles,  with  their  centers  on  the 
line  AB  extended.  The  equipotential  surfaces  are  circular 
cylinders,  which  are  shown  in  Fig.  47  as  circles  passing  through  the 
centers  A  and  B  of  the  conductors.  Within  the  conductors  them- 
selves there  are  no  equipotential  surfaces. 

For  purposes  of  analysis  it  is  convenient  to  regard  the  field  in 
Fig.  47  as  the  result  of  the  superposition  of  two  simpler  fields,  simi- 
lar to  those  of  the  concentric  cable  of  the  preceding  article.  Con- 


FIG.  47. — The  magnetic  field  produced  by  a  single-phase  transmission  line. 

sider  the  conductor  A,  together  with  a  concentric  cylinder  of  an 
infinitely  large  radius,  as  one  conducting  system.  Let  the  current 
flow  through  A  toward  the  reader,  and  return  through  the  infinite 
cylinder.  Let  the  conductor  B  with  a  similar  concentric  cylinder 
form  another  independent  system.  The  currents  in  the  conduc- 
tors A  and  B  are  to  be  the  same  as  the  actual  currents  flowing 
through  them,  but  each  infinite  cylinder  is  to  serve  as  a  return  for 
the  corresponding  conductor,  as  if  there  were  no  electrical  connec- 
tion between  A  and  B.  The  currents  in  the  two  cylinders  are 
flowing  in  opposite  directions  and  the  cylinders  themselves 


CHAP.  XI]     INDUCTANCE  OF  TRANSMISSION  LINES  195 

coincide  at  infinity,  because  the  distance  AB  between  their  axes  is 
infinitely  small  as  compared  to  their  radii.  Hence,  the  two  cur- 
rents in  the  cylinders  cancel  each  other,  and  the  combination  of 
the  two  component  systems  is  magnetically  identical  with  the 
given  loop. 

In  a  medium  of  constant  permeability,  the  resultant  magnetic 
intensity  H,  produced  at  a  point  by  the  combined  action  of  several 
independent  m.m.fs.,  is  equal  to  the  geometric  sum  of  the  intensi- 
ties produced  at  the  same  point  by  the  separate  m.m.fs.1  This 
being  true  of  the  intensities,  the  component  flux  densities  at  any 
point  are  also  combined  according  to  the  parallelogram  law 
because  they  are  proportional  to  the  intensities.  Hence,  the 
resultant  flux  can  be  regarded  as  the  result  of  the  superposition 
of  the  fluxes  created  by  the  component  systems. 

The  field  produced  by  the  system  A  consists  of  concentric  cir- 
cles, the  flux  density  outside  the  conductor  A  being  inversely  pro- 
portional to  the  distance  from  the  center  of  A  (curve  qr  in  Fig.  46). 
The  field  created  by  the  system  B  consists  of  similar  circles  around 
B,  and  the  field  shown  in  Fig.  47  is  a  superposition  of  these  two 
fields.  Thus  the  resultant  field  intensity  H  at  a  point  P  is  a 
geometric  sum  of 

Hi=i/2*ri, (115) 

and 

H2=i/2xr2, (116) 

HI  and  H2  being  perpendicular  to  the  corresponding  radii  vectors 
TI  and  r2  from  the  centers  of  the  conductors  to  the  point  P.  The 
directions  of  HI  and  H2  are  determined  by  the  right-hand  screw 
rule.  Since  HI  and  H2  are  known  in  magnitude  and  direction  at 
each  point  of  the  field,  the  resultant  intensity  H  may  also  be  deter- 
mined. 

To  deduce  the  equation  of  the  lines  of  force  in  the  resultant 
field,  we  shall  express  analytically  the  condition  that  the  total  flux 
which  crosses  the  surface  CP  is  equal  to  zero,  provided  that  C  and 
P  lie  on  the  same  line  of  force.  This  total  flux  may  be  considered 

1  This  principle  of  superposition  can  be  considered  (a)  as  an  experimental 
fact ;  (6)  as  an  immediate  consequence  of  the  fact  that  in  a  medium  of  constant 
permeability  the  effects  are  proportional  to  the  causes*;  (c)  as  a  consequence 
of  Laplace's  law  dH  =  Const. Xids  sin  0/10r2,  according  to  which  the  total 
field  intensity  is  regarded  as  the  sum  of  those  produced  by  the  infinitesimal 
elements  of  the  current,  or  currents. 


196  THE  MAGNETIC  CIRCUIT  [ART.  60 

as  the  resultant  of  the  fluxes  due  to  the  systems  A  and  B.  Accord- 
ing to  eq.  (115),  the  flux  density  due  to  A,  at  a  distance  x  from  A, 
is  BI  =  jj.i/2xx,  so  that  the  flux  due  to  A  which  crosses  CP  is 


=  C 

t/A 


(117) 


This  flux  is  directed  to  the  left,  looking  from  C  to  P.  By  analogy, 
the  flux  due  to  the  system  B  is 

02f  =  (fjLi/2n)'Ln(r2/BC)  maxwells/cm.,     .     .     .     (118) 

and  is  directed  to  the  right,  looking  from  C  to  P.  The  condition 
that  no  flux  crosses  CP  is,  that  </Y  is  equal  to  $2',  or 

Ln(r!/AC)=Ln(r2/£C), 
or 

ri/r2=AC/BC=Const  .....     (119) 

This  is  the  equation  of  a  line  of  force  in  "  bipolar  "  co-ordinates; 
the  curve  is  such  that  the  ratio  of  r*i  to  r2  remains  constant,  How- 
ever, this  constant  is  different  for  each  line  of  force,  because  each 
line  has  its  own  point  C. 

Eq.  (119)  may  be  proved  to  represent  a  circle,  by  selecting  an 
origin,  say  at  A,  and  substituting  for  rx  and  r2  their  values  in  terms 
of  the  rectangular  co-ordinates  x  and  y.  The  following  proof  by 
elementary  geometry  leads  to  the  same  result.  Produce  AP  and 
lay  off  PD=PB=r2.  According  to  eq.  (119),  BD  is  parallel  to 
CP,  and  consequently  PC  bisects  the  angle  APB  =to.  Let  the 
point  C'  lie  on  the  same  line  of  force  with  C;  then  no  flux  passes 
through  PC',  and  by  analogy  with  eq.  (119)  we  have 

n/r2=AC'/BC'=C<mat.        ....     (120) 

By  plotting  PD'  =r2  (not  shown  in  figure)  along  PA,  in  the  oppo- 
site direction  from  PD,  and  connecting  D'  to  B,  one  can  show  as 
before  that  PC'  bisects  the  angle  BPD  =  l8Q0-w.  But  the 
bisectors  of  two  supplementary  angles  are  perpendicular  to  each 
other;  consequently,  CPC'  is  a  right  angle,  and  the  point  P  lies 
on  a  semicircle  erected  on  the  diameter  CC'.  This  semicircle  is  the 
line  of  force  itself,  because  all  the  points,  such  as  P,  which  are  deter- 
mined by  C  and  C'  must  lie  on  it.  Another  semicircle  below  the 
line  AB  closes  the  line  of  force. 


CHAP.  XI]     INDUCTANCE  OF  TRANSMISSION  LINES  197 

From  eqs.  (119)  and  (120)  the  following  expressions  are  obtained 
for  the  radius  R  of  the  line  of  force  : 

BC 


or 

BC' 

•  ••••  •  •  •  (122) 


so  that  the  line  of  force  can  be  easily  drawn  for  a  given  C  or  C". 

To  prove  that  the  equipotential  lines  are  also  circles  we  proceed 
as  follows.  The  line  AB  is  evidently  an  equipotential  line,  because 
it  is  perpendicular  to  all  the  lines  of  force.  The  difference  of  mag- 
netic potent'al  or  the  m.m.f.  between  AB  and  P,  contributed  by 
the  system  A,  is  equal  to  i(d\/2n)  ampere-turns,  where  the  angle 
BAP  is  denoted  by  0\.  This  is  because  the  m.m.f.  due  to  the  sys- 
tem A,  taken  around  a  complete  circle  concentric  with  A,  is  equal 
to  i,  and  is  distributed  uniformly  along  the  circle,  for  reasons  of 
symmetry.  Or  else  it  follows  directly  from  eq.  (115).  By 
analogy,  the  difference  of  magnetic  potential  between  AB  and  P, 
due  to  the  system  B,  is  i(6i/lTi).  Thus  the  total  difference  of 
potential  between  AB  and  P  is 

Mcp  =  (i/^)(dl+d2)=(i/2n)(n-^.       .     .     (123) 

This  shows  that  the  m.m.f.  between  any  two  points  in  the  field  is 
proportional  to  the  difference  in  the  angles  CD  at  which  the  line  AB 
is  visible  from  these  points.  For  any  two  points  on  the  same 
equipotential  line  Mcp  is  the  same,  so  that  the  equation  of  such  a 
line  is 

&>=const  ........     (124) 

This  represents  the  arc  of  a  circle  passing  through  A  and  B,  and 
corresponding  to  the  inscribed  angle  a). 

Prob.  12.  A  single-phase  transmission  line  consists  of  two  conductors 
1  cm.  in  diameter,  and  spaced  100  cm.  between  the  centers.  Draw 
the  curve  of  flux  density  distribution  (pqst  in  Fig.  47)  for  an  instantaneous 
value  of  the  current  equal  to  100  amp. 

Ans.  z  =  50.0    25.0      0.5       -0.5  -50.0       -oo  cm.; 

B=  1.60    2.1380.40  -79.60  -0.53      0  maxw./sq.  cm. 
Prob.  13.    For  the  transmission  line  in  problem  12  draw  the  lines 


198  THE  MAGNETIC  CIRCUIT  [ART.  61 

of  force  which  divide  the  total  flux  between  the  wires  into  ten  equal 
parts  (not  counting  the  flux  within  the  wires). 

Ans.    The  circles  nearest  to  00'  cross  AB  at  a  distance  of  48.4 

cm.  from  each  other. 

Prob.  14.  Referring  to  the  two  preceding  problems,  draw  ten 
equipotential  circles  which  divide  the  whole  m.m.f.  of  100  ampere-turns 
into  ten  equal  parts. 

Ans.    The  arcs    nearest  to  AB  intersect  00'  at  a  distance  of 

32.5  cm.  from  each  other. 

Prob.  16.  A  telephone  line  runs  parallel  to  a  single-phase  power 
transmission  line.  The  position  of  one  of  the  telephone  wires  is  fixed; 
show  how  to  determine  the  position  of  the  other  wire  so  as  to  have  a 
minimum  of  inductive  disturbance  in  the  telephone  circuit.  Hint: 
The  center  lines  of  the  two  telephone  wires  must  intersect  the  same  line 
of  force  due  to  the  power  line. 

Prob.  16.  A  telephone  line  runs  parallel  to  a  25-cycle,  single-phase 
transmission  line.  The  distances  from  one  of  the  telephone  wires  to 
the  power  wires  are  3.5  m.  and  2.7  m.;  the  distances  from  the  other 
telephone  wire  to  the  power  wires  are  3.6  and  2.5  m.  (in  the  same  order). 
What  voltage  is  induced  in  the  telephone  line  per  kilometer  of  its  length, 
when  the  current  in  the  power  line  is  100  effective  amperes? 

NOTE:  In  practice,  this  voltage  is  neutralized  by  transposing 
either  line  after  a  certain  number  of  spans.  Ans.  0.33  volt. 

61.  The  Inductance  of  a  Single-phase  Line.  The  inductance  of 
a  single-phase  line  (Fig.  47)  can  be  calculated  according  to  the  fun- 
damental formulae  (105)  or  (106),  provided  that  the  permeances 
of  the  elementary  paths  be  expressed  analytically.  However,  in 
this  case  it  is  much  simpler  to  use  the  principle  of  superposition 
employed  in  the  preceding  article,  and  to  consider  the  actual  flux 
as  the  resultant  of  two  fluxes  each  surrounding  concentrically  one 
of  the  wires  and  extending  to  infinity.  The  fluxes  produced  by 
the  two  component  systems  are  equal  and  symmetrical  with 
respect  to  the  wires.  It  is  therefore  sufficient  to  calculate  the 
linkages  of  the  loop  AB  with  the  flux  produced  by  one  of  the  sys- 
tems, say  that  corresponding  to  A,  and  to  multiply  the  result  by 
two. 

The  flux  produced  by  A  and  having  A  as  a  center  link,  with 
the  current  in  the  loop  AB.  These  linkages  may  be  divided  into 
the  following: 

(a)  Linkages  within  the  wire  A ;  that  is,  from  x  =0  to  x  =a; 

(6)  Linkages  between  the  wires  A  and  B,  that  is,  from 

x=a  to  x  =b—  a; 


CHAP.  XIJ     INDUCTANCE  OF  TRANSMISSION  LINES  199 

(c)  Linkages  outside  the  loop,  that  is,  from  x=b+a  to  x  = 
infinity, 

(d)  Linkages  within  the  wire  B,  that  is,  from 

^. 

x=b—  a  to  x=b+a. 

The  linkages  (a),  (b)  and  (c)  are  the  same  as  in  a  concentric 
cable  (Fig.  46),  because  the  shape  of  the  lines  of  force  and  the 
number  of  turns  with  which  they  are  linked  are  the  same.  The 
partial  linkages  (d)  are  somewhat  difficult  to  express  analytically. 
When  the  distance  b  between  the  wires  is  large  as  compared  to 
their  diameters,  the  whole  current  in  B  may  be  assumed  to  be  con- 
centrated along  the  axis  of  the  wire  B,  instead  of  being  spread  over 
the  cross-section.  With  this  assumption,  the  partial  linkages  (d) 
are  done  away  with,  the  linkages  (6)  are  extended  to  x  =b,  and  the 
linkages  (c)  begin  from  x  =b.  The  expressions  for  the  linkages  (a) 
and  (b)  are  given  by  eqs.  (110)  and  (109)  respectively.  The  link- 
ages (c)  are  equal  to  zero,  because  in  this  region  the  lines  of  force 
produced  by  A  are  linked  with  both  A  and  B,  and  therefore 
with  i—  i=Q  ampere -turns.  Thus,  the  inductance  in  question  is 

L'  =0.46  logio(6/o)  +0.05 (125) 

This  gives  the  inductance  of  a  single-phase  line  in  perms  per  centi- 
meter length,  or  in  millihenry s  per  kilometer  length  of  the  wire.1 
To  obtain  the  inductance  per  unit  length  of  the  line  this  expression 
must  be  multiplied  by  two,  because  the  linkages  due  to  the  flux 
produced  by  the  system  B  are  not  taken  into  account  in  eq.  (125). 
However,  for  the  purposes  of  the  next  two  articles  it  is  more  con- 
venient to  use  expression  (125),  and  to  consider  separately  the 
inductance  of  each  wire,  remembering  that  the  two  wires  of  a  loop 
are  in  series,  and  that  therefore  their  inductances  are  added.2 

Prob.  17.  Check  by  means  of  formula  (125)  some  of  the  values  for 
the  inductance  and  reactance  of  transmission  lines  tabulated  in  the 
various  pocketbooks  and  handbooks. 

1  It  is  of  interest  to  note  that  the  exact  integration  over  the  partial  linkages 
(d)  leads  to  the  same  Eq.  (125),  so  that  this  formula  is  correct  even  when 
the  wires  are  close  to  each  other.     See  A.  Russell,  Alternating  Currents,  Vol. 
1  (1904),  pp.  59-60. 

2  The  inductance  of  two  or  more  parallel  cylinders  oY  any  cross-section  can 
be  expressed  through  the  so-called  "  geometric  mean  distance,"  introduced 
by  Maxwell.    For  details  see  Orlich,   Kapazitdt  und  Induktivitat  (1909), 
pp.  63-74. 


200  THE  MAGNETIC  CIRCUIT  [ART.  61 

Prob.  18.  Show  by  means  of  tables  or  curves  that  the  inductance 
of  a  transmission  line  varies  much  more  slowly  than  (a)  the  spacing 
with  a  given  size  of  wire,  (6)  the  size  of  wire  with  a  given  spacing. 

Prob.  19.  When  the  diameters  of  the  two  wires  A  and  B  are  different, 
prove  that  the  inductance  of  the  complete  loop  is  the  same  as  if  the 
diameter  of  each  wire  was  equal  to  the  geometric  mean  of  the  actual 
diameters. 

Prob.  20.  Show  that  the  inductance  of  a  single-phase  line  with  a 
ground  return  can  be  calculated  from  eq.  (125)  by  putting  b=2h  where 
h  is  the  elevation  of  the  wire  above  the  ground.  Hint:  In  Fig.  47  the 
plane  00'  may  be  considered  to  be  the  surface  of  the  earth,  assumed 
to  be  a  perfect  conductor.  If  the  earth  be  removed,  an  "image"  con- 
ductor B  must  be  added  in  order  to  provide  a  return  path  for  the 
current,  such  that  the  field  surrounding  A  would  remain  the  same. 

Prob.  21.  Two  single-phase  lines  are  placed  on  two  cross-arms  of 
the  same  tower,  one  directly  above  the  other,  at  a  vertical  distance 
of  c  cm  apart.  What  is  the  total  inductance  of  the  combination,  when 
the  two  lines  are  connected  in  parallel  and  each  line  carries  one-half 
of  the  total  current? 

Solution:  Consider  the  four  wires  as  forming  four  fictitious  systems, 
with  cylinders  at  infinity  as  returns.  Let  b  be  the  spacing  in  each 
loop,  and  let  b  be  larger  than  c.  Denote  the  wires  in  one  loop  by  1 
and  2,  in  the  other  by  1'  and  2',  and  let  d  be  the  diagonal  distance 
between  1  and  2'.  Assume  all  the  wires  except  1  to  be  of  an  infinitesimal 
cross-section.  Then,  the  linkages  of  the  flux  produced  by  the  system 
1  with  the  currents  in  the  four  wires  are 


Ln(c/o)  +0.2i 

+0.2(ii)2  Ln(d/6)  milli  joules  /km. 

Thus,  allowing  the  same  amount  for  the  linkages  due  to  the  cur- 
rent in  the  wire  1',  we  get  that  the  inductance  of  the  split  line,  each  way, 
is 

(bd/ca)  +0.05]  millihenrys/km., 


instead  of  the  expression  (125)  for  the  single  line.  The  same  result 
is  obtained  when  b  is  smaller  than  c.  Hence,  by  splitting  a  line  in 
two  the  inductance  is  considerably  reduced,  because  partial  linkages  are 
substituted  for  some  of  the  complete  linkages.  If  d  were  equal  to  c 
the  reduction  would  be  50  per  cent;  but  since  d  is  always  larger  than 
c  the  gain  is  less  than  50  per  cent.  However,  when  the  two  lines  are 
very  far  apart  the  saving  is  very  nearly  50  per  cent. 

Prob.  22.  A  certain  single-phase  transmission  line  has  been  designed 
to  consist  of  No.  000  B.  &  S.  conductor  with  a  spacing  of  180  cm.  It 
is  desired  to  reduce  the  reactive  drop  by  about  20  per  cent,  without 
increasing  the  weight  of  copper,  by  using  two  lines  in  parallel,  with 
the  same  spacing.  What  is  the  size  of  the  conductor  and  the  distance 
between  the  loops?  Ans.  No.  1  B.  &  S.  ;  about  8  cm. 

Prob.  23.     Solve  problem  21    when  the  load  is  divided  unequally 


CHAP.  XI]     INDUCTANCE  OF  TRANSMISSION  LINES  201 

between  the  two  loops,  the  currents  being  mi  and  ni  respectively,  where 


Ans.    L'  =  0.46[(ra2  +  tt2)    log  (c/a)  +log  (6/c)  +2mn  log 

0.05(ra2+n2),  when  6>c,  and  L'=0.46[(ra2+n2)  log  (6  /a)  + 
2ranlog  (d/c)]+0.05(w2+n2),  when  b<c. 

Prob.  24.  A  single-phase  line  consists  of  three  conductors,  the 
total  current  flowing  through  conductor  1,  and  returning  through  con- 
ductors 2  and  3  in  parallel.  If  the  current  in  one  return  conductor  is 
mi,  and  in  the  other  ni,  where  m+n  =  l,  prove  that  the  inductance  of 
the  line  per  kilometer  of  its  length  is,  in  millihenrys, 

L'  =  0.46  [log  (ftu/oO  +w2  log  (623/<z2)  +n2  log  (623/a3)  +m  log  (bia/618) 

(&12/623)  +nlog  (&18/b28)]  +0.05(1  +m2+n2),  when  bn>bl3>b23. 


In  the  particular  case  when  bl2  =  b23  =  bn,  al=a2  =  a3,  and  m=n  =  %,  the 
inductance  is  reduced  by  25  per  cent  as  compared  to  that  of  a  single  loop.1 

62.  The  Inductance  of  a  Three-phase  Line  with  Symmetrical 
and  Semi-symmetrical  Spacing.  The  magnetic  field  which  sur- 
rounds a  single-phase  line  varies  in  its  intensity  from  instant  to 
instant,  as  the  current  changes,  but  the  direction  of  the  magnetic 
intensity  and  of  the  flux  density  at  each  point  remains  the  same. 
In  other  words,  the  flux  is  a  pulsating  one.  The  field  created  by 
three-phase  currents  in  a  transmission  line  varies  at  each  point  in 
both  its  magnitude  and  direction.  At  the  end  of  each  cycle,  the 
field  assumes  its  original  magnitude  and  direction.  If  the  spacing 
of  the  wires  is  symmetrical,  the  field  at  the  end  of  each  third  of  a 
cycle  has  the  same  magnitude  and  position  with  respect  the  next 
wire.  The  field  may  therefore  be  said  to  be  revolving  in  space. 

This  revolving  flux,  like  that  in  an  induction  motor,  induces 
e.m.fs.  in  the  three  phases.  The  problem  is  to  determine  these 
counter-e.m.fs.  in  the  transmission  line,  knowing  the  size  of  the 
wires,  the  spacing,  and  the  load.  In  transmission  line  calculations, 
especially  in  determining  the  voltage  drop  and  regulation,  it  is 
convenient  to  consider  each  wire  separately,  and  to  determine  the 
voltage  drop  in  phase  and  in  quadrature  with  the  current.  Thus, 
having  expressed  the  e.m.fs.  induced  by  the  revolving  flux  in  terms 
of  the  constants  of  the  line,  each  wire  is  then  considered  as  if  it 
were  brought  outside  the  inductive  action  of  the  two  other  wires. 

We  shall  consider  first  the  case  of  an  equidistant  spacing  of  the 
three  wires,  because  in  most  practical  calculations  of  voltage  drop 

1The  splitting  of  conductors  discussed  in  problems  21  to  24  has  been 
proposed  for  extremely  long  transmission  lines,  in  order  to  reduce  their  induct- 
ance and  at  the  same  time  increase  their  electrostatic  permittance  (capacity). 


202  THE  MAGNETIC  CIRCUIT  [ART.  62 

an  unsymmetrical  spacing  is  replaced  by  an  equivalent  equidistant 
spacing.  The  exact  solution  for  an  unsymmetrical  spacing  is  given 
in  the  next  article.  Let  the  instantaneous  values  of  the  three  cur- 
rents in  the  wires  A,  B  and  C  of  a  three-phase  transmission  line  be 
ii,  12  and  13.  The  sum  of  the  three  currents  at  each  instant  is 
zero,  or 

t'i +12+^3=0 (126) 

Let  4>eq  be  the  equivalent  flux  which  links  at  any  instant  with  the 
wire  A.  The  instantaneous  e.m.f.  induced  in  this  wire  is 

e1  =  -d#eq/dt.    . (127) 

The  equivalent  flux  consists  of  the  actual  flux  outside  the  wire  plus 
the  sum  of  the  fluxes  inside  the  wire,  each  infinitesimal  tube  of  flues 
being  reduced  in  the  proper  ratio,  according  to  the  fraction  of  the 
cross-section  of  the  wire  with  which  it  is  linked.  Or,  what  is  the 
same  thing,  each  wire  is  replaced  by  an  equivalent  hollow  cylinder 
of  infinitesimal  thickness,  without  partial  linkages,  as  in  problem 
9  in  Art.  59  (consult  also  the  definition  of  equivalent  permeance 
given  in  Art.  58). 

In  order  to  determine  4>eq  we  replace  the  three-phase  system 
by  two  superimposed  single-phase  systems.  The  current  i\  in  the 
wire  A  may  be  thought  of  as  the  sum  of  the  currents  —12  and  —13, 
each  flowing  in  a  separate  fictitious  wire,  and  both  of  these  wires 
coinciding  with  A.  The  currents  +12  in  B  and  — 12  in  A  form 
one  single-phase  loop,  while  the  currents  +t'3  in  C  and  —  i3  in  A 
form  the  other  loop.  The  flux  ®eq  which  surrounds  A  is  the  sum 
of  the  fluxes  produced  by  these  two  loops.  The  flux  per  unit 
length  of  the  line,  due  to  the  first  loop,  is  equal  to  —(Peqi2,  since 
the  number  of  turns  is  equal  to  one.  For  the  same  reason  (P'eq  = 
U  where  U  is  determined  by  eq.  (125).  Hence,  the  flux  per 
unit  length  of  the  line,  due  to  the  first  loop,  is  —  L'i2.  Similarly, 
the  flux  due  to  the  second  loop  and  linked  with  A  is  equal  to 
—L'iz,  the  same  value  of  U  being  used  because  the  spacing  and 
the  sizes  of  all  of  the  wires  are  the  same.  Thus, 


'eq 


(128) 


the  last  result  being  obtained  by  substituting  the  value  of  12+^3 
from  eq.  (126).     Thus,  eq.  (127)  becomes  simply 

ei  =  -L'dii/dt,  (129) 


CHAP.  XI]      INDUCTANCE  OF  TRANSMISSION  LINES  203 

that  is,  the  induced  e.m.f.  is  the  same  as  in  a  single-phase  line  carry- 
ing the  current  i\.  Thus,  the  inductance  of  a  three-phase  line  with 
symmetrical  spacing,  per  wire,  is  the  same  as  the  inductance  of  a 
single-phase  line,  per  wire,  with  the  same  size  of  wire  and  the  same 
spacing.  The  total  e.m.f.  induced  in  each  wire  is  in  quadrature 
with  the  current  in  the  wire. 

In  reaching  this  conclusion  the  following  facts  were  made  use 
of :  (a)  The  current  in  each  wire  at  any  instant  is  equal  to  the  sum 
of  the  currents  in  the  two  other  wires;  (6)  the  fluxes  due  to  sepa- 
rate m.m.fs.  can  be  superimposed  in  a  medium  of  constant  perme- 
ability; (c)  The  inductance  of  the  loop  A-B  is  equal  to  that  of  A-C 
because  of  the  same  spacing.  No  other  suppositions  in  regard  to 
the  character  of  the  load  or  the  voltages  between  the  wires  were 
made.  Therefore,  the  conclusion  arrived  at  holds  true: 

(a)  For  balanced  as  well  as  unbalanced  loads; 

(6)  For  balanced  or  unbalanced  line  voltages; 

(c)  For  a  three-wire    two-phase    system,   three-wire   single- 
phase  system,  monocyclic  system,  etc. 

(d)  For  sinusoidal  voltages  as  well  as  for  those  departing  from 
this  form. 

Semi-symmetrical  Spacing.  When  two  out  of  the  three  distances 
between  the  wires  in  a  three-phase  line  are  equal  to  each  other,  the 
arrangement  is  called  semi-symmetrical.  Two  common  cases  of 
this  kind  are :  (a)  When  the  wires  are  placed  at  the  vertices  of  an 
isosceles  triangle;  (6)  when  they  are  placed  at  equal  distances  in 
the  same  plane,  for  instance  on  the  same  cross-arm,  or  are  fastened 
to  suspension  insulators,  one  above  the  other.  In  such  cases  the 
inductive  drop  in  the  symmetrically  situated  wire  is  the  same  as  if 
the  wire  belonged  to  a  single-phase  loop,  carrying  the  same  current, 
and  with  a  spacing  equal  to  the  distance  of  this  wire  to  either  of 
the  other  two  wires.  Let,  for  instance,  the  distance  A-B  be  equal 
to  B-C,  and  let  the  distance  A-C  be  different  from  the  two.  The 
proof  given  above  can  be  repeated  for  the  wire  B,  and  the  same 
conclusion  will  be  reached  because  the  spacing  A-C  is  not  used 
in  the  deduction.  But,  of  course,  the  proof  does  not  hold  true  for 
either  wire  A  or  C. 

When  the  three  wires  are  in  the  same  plane,^  the  inductance  of 
each  of  the  outside  wires  is  larger  than  that  of  the  middle  wire. 
This  can  be  shown  as  follows :  Let  the  three  wires  be  in  a  horizontal 
plane,  and  let  them  be  denoted  from  left  to  right  by  A,  B,  C.  Let 


204  THE  MAGNETIC  CIRCUIT  [ART.  62 

the  distance  between  A  and  B  be  equal  to  b  and  the  distance 
between  A  and  C  be  equal  to  26.  If  the  wire  B  were  moved  to 
coincide  with  C,  the  inductance  of  A  would  be  the  same  as  if  it 
belonged  to  a  single-phase  loop  with  a  spacing  26.  If  C  were 
moved  to  coincide  with  B,  the  inductance  of  A  would  be  that  of  a 
wire  in  a  single-phase  loop  with  a  spacing  b.  Thus,  the  inductance 
of  A  corresponds  in  reality  to  a  spacing  intermediate  between  b  and 
26.  The  inductance  of  the  middle  wire  B  is  the  same  as  that  of  a 
wire  in  a  single-phase  loop  with  the  spacing  6,  as  is  proved  above. 
Thus,  the  inductance  of  either  A  or  C  is  larger  than  that  of  B. 

An  inspection  of  a  table  of  the  inductances  or  reactances  of 
transmission  lines  will  show  that  the  inductance  increases  much 
more  slowly  than  the  spacing.  For  instance,  according  to  the 
Standard  Handbook,  the  reactance  per  mile  of  No.  0000  wire,  at 
25  cycles,  is  0.303  ohm  with  a  spacing  of  72  inch,  and  is  0.340  ohm 
with  a  spacing  of  150  inch.  Therefore,  in  practical  calculations, 
when  the  spacing  is  semi-symmetrical,  the  values  of  inductance  are 
taken  the  same  for  all  the  three  wires,  for  an  average  spacing 
between  the  three,  or,  in  order  to  be  on  the  safe  side,  for  the  maxi- 
mum spacing.  In  the  most  unfavorable  case,  even  if  an  error  of 
say  10  per  cent  be  made  in  the  estimated  value  of  the  inductance, 
and  if  the  inductive  drop  is  say  20  per  cent  of  the  load  voltage,  the 
error  in  the  calculated  value  of  voltage  drop  is  only  2  per  cent  of 
the  load  voltage,  and  that  at  zero  power  factor.  At  power  factors 
nearer  unity,  when  the  vector  of  the  inductive  drop  is  added  at 
an  angle  to  the  line  voltage,  the  error  is  much  smaller. 

Prob.  25.  Show  that  the  instantaneous  electromagnetic  energy 
stored  per  kilometer  of  a  three-phase  line  with  symmetrical  spacing  is 
equal  to  iZ/fo'+^'+tV)  milli joules  per  kilometer,  where  U  has  the 
value  given  by  eq.  (125).  If  this  is  true,  then  each  wire  may  be  con- 
sidered as  if  it  were  subjected  to  no  inductive  action  from  the  other 
wires  and  had  an  inductance  L'  expressed  by  eq.  (125) .  This  is  another 
way  of  proving  eq.  (129),  and  the  statement  printed  in  italics  above. 
Solution:  Consider  each  wire,  with  a  concentric  cylinder  at  infinity, 
as  a  component  system.  Determine  the  linkages  of  the  field  created 
by  the  system  A  with  the  currents  in  A,  B,  and  C,  as  in  Art.  61.  The 
result  is  equal  to  \L'i?.  Similar  expressions  are  then  written  by  analogy 
for  the  fluxes  due  to  the  systems  B  and  C. 

Prob.  26.  Show  graphically  that,  when  the  distances  A-B  and 
A-C  are  equal,  the  equivalent  flux  linking  with  A  is  independent  of 
the  spacing  B-C,  and  is  the  same  as  if  B  and  C  coincided.  That  is, 
prove  that  the  inductance  of  A  is  the  same  as  if  it  belonged  to  a  single- 


CHAP.  XI]      INDUCTANCE  OF  TRANSMISSION  LINES 


205 


phase  loop.  Solution:  Let  7t,  72,  and  73  (Fig.  48),  represent  the  vectors 
of  the  three  currents  at  an  unbalanced  load.  The  current  /x  in  A  is 
replaced  by  — 12  and  —  73,  and  the  system  is  split  into  two  single-phase 
loops,  A-B  and  A-C.  The  fluxes  due  to  these  systems  and  linked  with 
A  are  denoted  by  fl>n  and  013.  They  are  in  phase  with  the  corresponding 
currents,  and  are  proportional  to  the  magnitudes  of  these  currents, 
because  of  the  equal  spacing.  Hence,  the  triangles  of  the  currents  and 
of  the  fluxes  are  similar,  and  the  resultant  flux  ^l  linking  with  A  is 
in  phase  with  7t.  If  0i2  =  £(P72,  and  i03  =  i(P73,  where  (P  is  the  equivalent 
permeance  of  each  single-phase  loop,  then  the  result  shows  that  (^i  =  ^(P71. 
If  the  wires  B  and  C  coincided  the  equivalent  permeance  (P  would  be 
the  same,  and  hence  the  proposition  is  proved.  The  voltage  drop,  Eit 
due  to  the  flux  01  is  shown  by  a  vector  leading  /i  by  90  degrees. 

63.  The  Equivalent  Reactance  and  Resistance  of  a  Three- 
phase  Line  with  an  Unequal  Spacing  of  the  Wires.  In  the  case 
of  an  unequal  spacing  of  the  wires  eqs.  (126)  and  (127)  still  hold 
true,  because  they  do  not  depend  upon  the  spacing;  but  eq.  (128) 
becomes 

./..     (130) 


-i 


where  Z/i2  is  the  value  of  the  inductance  per  unit  length,  calcu- 
lated by  eq.  (125)  for  the  spacing  between  A  and  B,  and  L'i3  is 
the  value  of  the  inductance  per  unit 
length,  for  the  spacing  A-C.     Substi-  Ti 

tuting  the  valueof  4>eq  from  eq.  (130) 
into  eq.  (127)  we  get 

This  shows  that  with  an  unequal 
spacing  the  effect  of  the  mutual 
induction  of  the  phases  cannot  be 
replaced  by  an  equivalent  inductance 
in  each  phase,  because,  generally 
speaking,  the  currents  12  and  i3  cannot 
be  eliminated  from  this  equation  by 
means  of  eq.  (126). 

Let  us   apply    now  eq.    (131)    to 
the   case  of    sinusoidal    currents  and 

voltages.  Let  the  current  in  the  wire  B  be  12  =  v2l 2  sin  Inft, 
where  72  is  the  effective  value  of  the  current;  then  the  first  term 
on  the  right-hand  side  of  eq.  (131)  becomes  27r/7/i2V272  cos  2-ft. 


FIG.  48.— The  currents  and 
fluxes  in  a  three-phase  line 
with  a  symmetrical  spacing. 


206  THE  MAGNETIC  CIRCUIT  [ART.  63 

In  the  symbolic  notation  this  is  represented  as  jx^'l,  where  #12' 
=27r/Li2/  is  the  reactance  corresponding  to  1/12',  1  2  is  the  vector 
of  the  effective  value  of  the  current  in  the  wire  B,  and  j  signifies 
that  the  vector  x\z\2  is  in  leading  quadrature  with  the  vector  12. 
Consequently,  the  voltage  drop  E\  in  the  wire  A,  equal  and 
opposite  to  the  induced  e.m.f  .,  is 

Ei  =  -3Xi2'l2-jxi3'Iz  ......     (132) 

When  the  currents  are  given,  /2  and  73  can  be  expressed  in  the 
usual  way  through  their  components,  and  the  drop  EI  is  then 
expressed  through  its  components  as  e\+jef\.  The  reactances 
x\2  and  £13'  are  taken  from  the  available  tables,  for  the  specified 
frequency  and  the  appropriate  spacings,  or  else  they  can  be 
calculated  using  the  value  of  L'  from  eq.  (125). 

The  voltage  drop  E±  in  eq.  (132)  can  be  represented  as  if  it  were 
due  to  an  equivalent  reactance  x\  and  an  equivalent  resistance  r\ 
in  the  phase  A  (the  latter  in  addition  to  the  actual  ohmic  resistance 
of  the  wire)  .  This  is  possible  when  1  2  and  1  3  can  be  expressed  in 
terms  of  /i,  and  is  especially  convenient  whenever  the  phase  differ- 
ence between  these  currents  and  their  ratio  is  constant.  Namely, 
let  the  current  I2  lead  the  current  l\  in  phase  by  <£i2  electrical 
degrees  (Fig.  49)*.  Then 

.    .     .     (133) 


where  (/2//i)  is  the  ratio  of  the  effective  values  of  the  currents, 
apart  from  their  phase  relation.  Multiplying  the  vector  /i  by 
(/2//i)  changes  its  magnitude  to  that  of  72,  while  multiplying  it  by 
(cos  (£12  +  7  sin  $12)  turns  it  counter-clockwise  by  </>i2  degrees.  By 
analogy  we  also  have  that 

/3  =  (VW{i(cos<£i3+/sin<£i3).     .     .     .     (134) 

Both  (£12  and  <£i3  are  measured  counter-clockwise.  Substituting 
these  values  into  eq.  (132)  and  separating  the  real  from  the 
imaginary  part  we  get 

EI  =Ii[(l2/h)xi2    sin  fa  +  Vs/Idxis'  sin  <£i3] 

-H\[(h/h)x\*  cos  <f>i2  +  (h/h)xiz'  cos  018].  .     .     (135) 

Thus,  the  drop  El  is  the  same  as  if  it  were  caused  by  a  fictitious 
reactance 

Xi    =  -  (l2/Il)Xi2    COS  $12  -  (/3//l)Zl3'  COS  <£i3,      .       (136) 


CHAP.  XI]      INDUCTANCE  OF  TRANSMISSION  LINES 


207 


and  a  fictitious  resistance 


sn 


sn 


(137) 


-i 


Both  x\  and  r\  may  be  either  positive  or  negative,  depending 
upon  the  constants  of  the  circuit  and  of  the  load.  The  resistance 
TI  does  not  involve  any  loss  of 
power,  converted  into  heat;  it 
merely  shows  that  energy  is  trans- 
ferred inductively  from  phase  A 
into  B  or  C,  at  a  rate  I-pr\,  due  to 
a  lack  of  symmetry  in  the  resultant 
field. 

Prob.    27.        Show  that   with    a 
balanced  three-phase  load 


0.866  (z12'  - 


Prob.  28.  When  the  three  wires 
are  in  the  same  plane,  .the  spacings 
being  equal,  and  the  three-phase  load 
balanced,  show  that  the  equivalent  reactance  of  each  outside  wire 


FIG.  49.  —  The  currents  and  fluxes 
m  a  three-phase  line  with  an 
unsymmetriccd  spacing. 


s</=*m'+0.435/XlO-3  ohm/km., 


(139) 


where  Xm  is  the  reactance  of  the  middle  wire  per  kilometer,  in  ohms. 
The  equivalent  resistance  of  the  middle  wire  is  zero,  and  that  of  the 
two  outside  wires  is 


r0'  =  ±0.753/X10-3  ohm/km., 


(140) 


where  the  sign  plus  refers  to  the  wire  in  which  the  current  leads  that  in 
the  middle  wire. 

Prob.  29.  Compare  the  vector  diagram  in  Fig.  49  with  that  in  Fig. 
48,  and  shown  that  with  an  unsymmetrical  spacing  the  induced  voltage 
E!  is  not  in  quadrature  with  the  corresponding  current  Ilt  so  that  the 
action  of  the  other  two  wires  cannot  be  replaced  by  an  equivalent 
inductance  alone,  but  only  by  an  inductance  and  a  resistance.  Show 
graphically  that  the  latter  may  be  either  positive  or  negative. 


CHAPTER  XII 

THE    INDUCTANCE    OF    THE    WINDINGS    OF 
ELECTRICAL    MACHINERY. 

64.  The  Inductance  of  Transformer  Windings.  When  a 
transformer  is  operated  at  no  load,  i.e.,  with  its  secondary  cir- 
cuit open,  practically  the  whole  flux  is  concentrated  within 
the  iron  core.  When,  however,  the  transformer  is  loaded,  so  that 
considerable  currents  flow  in  both  windings,  appreciable  leakage 
fluxes  are  formed  (Fig.  50),  which  are  linked  partly  with  the 
primary  winding,  and  partly  with  the  secondary  winding.  When 
the  load  current  is  considerable,  the  primary  and  the  secondary 
ampere-turns  are  large  as  compared  to  the  exciting  ampere- 
turns,  so  that  at  each  instant  the  secondary  ampere-turns  are 
practically  equal  and  opposite  to  the  primary  ampere-turns. 
An  inspection  of  Fig.  50  will  show  that  the  m.m.f.  acting  upon 
the  useful  path  in  the  iron  is  equal  to  the  difference  between  the 
m.m.fs.  of  the  primary  and  secondary  windings,  while  the  m.m.f. 
acting  upon  leakage  paths  is  equal  to  the  sum  of  the  m.m.fs. 
of  both  windings. 

Take,  for  instance,  the  line  of  force  fghk ;  with  respect  to  the 
part  fg  of  its  path,  the  secondary  coil  Si  and  the  adjacent  half 
PI  of  the  primary  coil  form  together  a  fictitious  annular  coil 
(leakage  coil).  The  m.m.f.  of  this  coil  is  equal  to  in^,  where 
ii  is  the  primary  current,  and  n^  is  the  number  of  turns  in  the 
whole  primary  coil  P.  Similarly,  the  coil  S2  and  the  part  P% 
of  the  primary  coil  may  be  said  to  form  another  fictitious  coil 
linking  with  the  part  hk  of  the  path  of  the  lines  of  force. 

It  will  be  seen  from  the  dots  and  crosses  that  the  m.m.fs. 
of  the  two  fictitious  coils  assist  each  other,  and  that  the  paths 
of  the  leakage  flux  are  as  indicated  by  the  arrow-heads.  Some 
lines  of  force  are  linked  with  the  total  m.m.f.  of  the  fictitious 
coils,  others  are  linked  with  only  part  of  the  turns.  Although 

208 


CHAP.  XII] 


INDUCTANCE  OF  WINDINGS 


209 


the  reluctance  of  the  leakage  paths  is  very  high  as  compared 
to  that  of  the  useful  path  in  iron,  yet  the  m.m.f.  acting  upon 


.Actual  Field 


Simplified  Fiel 


Flux  Density  Distribution 


X 
X 

Jr 

<e£G-»j 

?v2 

""ir 

X 

^ 

FIG.  50. — The  leakage  field  in  a  transformer  with  cylindrical  coils. 

the  leakage  paths  is  also  many  times  greater  than  that  acting 
upon  the  path  in  iron.  As  a  consequence,  the  leakage  fluxes 
reach  appreciable  magnitudes. 


210  THE   MAGNETIC  CIRCUIT  [ART.  64 

The  leakage  fluxes  induce  e.m.fs.  in  the  windings  in  lagging 
quadrature  with  the  respective  currents,  and  thus  affect  the 
voltage  regulation  of  the  transformer.  That  part  of  the  applied 
voltage  which  balances  these  e.m.fs.  is  known  as.  the  reactance 
drop  in  the  transformer.  It  is  customary  to  speak  about  the 
primary  reactance  and  the  secondary  reactance,  also  about  the 
primary  and  the  secondary  leakage  fluxes,  as  if  they  had  a 
separate  and  independent  existence.  However,  it  must  be  under- 
stood that  each  leakage  flux  is  produced  by  the  combined  action 
of  both  windings,  as  is  explained  above.  Moreover,  where  the 
leakage  fluxes  enter  the  iron  they  combine  with  the  main  flux 
in  the  proper  direction,  so  that  they  form  there  only  a  com- 
ponent of  the  resultant  flux. 

In  reality,  the  primary  ampere-turns  are  not  exactly  equal 
and  opposite  to  the  secondary  ampere-turns,  so  that,  in  addition 
to  the  leakage  fluxes  shown  in  Fig.  50,  there  is  a  leakage  flux 
due  to  the  magnetizing  ampere-turns.  However,  this  correction 
is  negligible,  when  the  load  is  considerable,  and  the  calculation 
of  the  leakage  flux  is  greatly  simplified  by  assuming  the  primary 
ampere-turns  to  be  exactly  equal  and  opposite  to  the  secondary 
ampere-turns. 

The  effect  of  the  leakage  reactance  upon  the  performance 
of  a  transformer  is  treated  in  The  Electric  Circuit;  there  the 
value  of  the  reactance  is  supposed  to  be  given.  Here  the  problem 
is  to  show  how  to  calculate  the  leakage  inductance  from  the  given 
dimensions  of  a  transformer.  The  two  types  of  winding  to  be 
considered  are  the  one  with  cylindrical  coils  (Fig.  50)  and  the 
one  with  flat  coils  (Fig.51).  Both  types  of  winding  can  be  used 
with  any  of  the  three  kinds  of  magnetic  circuit  which  are  used 
with  transformers  (Figs.  12,  13,  and  14). 

The  problem  of  calculating  the  leakage  inductance,  according 
to  the  fundamental  formula  (106),  is  reduced  to  that  of  finding 
the  permeances  of  the  individual  paths  of  the  leakage  flux.  It 
would  be  out  of  the  question  here  to  determine  the  actual  paths 
and  to  express  their  permeances  mathematically.  Therefore,  in 
accordance  with  Dr.  Kapp's  proposal,1  simplified  paths  are 
assumed,  shown  in  Fig.  50  to  the  right.  The  inductance  so 
calculated  is  corrected  by  an  empirical  factor,  obtained  from 
experiments  on  transformers  of  similar  type  and  proportions. 

1  G.  Kapp,  Transformers  (1908),  p.  177. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  211 

The  simplifying  assumptions  are  (a)  that  the  paths  within  and 
between  the  coils  are  straight  lines,  and  (6)  that  the  reluctance 
of  the  paths  in  the  space  outside  the  coils  is  negligible,  because 
the  cross-section  of  these  paths  is  practically  unlimited. 

(1)  Cylindrical  Coils.  We  shall  calculate  first  the  primary 
inductance  of  a  transformer  having  cylindrical  coils,  i.e.,  the 
inductance  due  to  the  linkages  of  the  leakage  flux  with  the 
primary  winding.  The  permeance  of  the  path  of  the  complete 
linkages  is  (Pci  =  J*.a<i0m/2l  perms,  where  Om  is  the  mean  length 
of  a  turn  in  the  coil  P,  and  0,1  is  the  radial  thickness  of  the  flux. 
The  notation  is  shown  in  the  detail  drawing,  at  the  bottom  of  Fig. 
50.  In  this  expression  a,iOm  is  the  mean  cross-section  of  the 
path,  being  an  average  between  the  cross-sections  within  the 
spaces  PI~SI  and  P2-S2.  The  length  of  the  paths  within  the 
coils  is  21,  and  the  reluctance  of  the  paths  outside  the  coils  is 
neglected.  This  path  is  linked  with  n^  turns. 

Similarly,  the  permeance  of  an  infinitesimal  annular  path 
within  the  primary  coil,  at  a  distance  x  from  its  center,  and 
of  a  width  dx,  is  d(PPi  =  jj.0mdx/2l  perms.  This  path  is  linked 
with  npi  =  ni(2x/bi)  turns.  Substituting  these  values  into  eq. 
(106)  we  obtain 


a, 

or 

L1  =  (fJLU12Om/2l)(a1  +  ib1)  perms.     .     *     «'    .     (141) 

By  a  somewhat  similar  reasoning  we  should  find  for  the  com- 
bination of  the  two  secondary  coils,  assuming  them  to  be  connected 
electrically  in  series, 

L2=(/m22Ow/2Z)(a2  +  i&2)  perms.     ....     (142) 

In  the  operation  of  a  transformer  it  is  the  total  equivalent 
inductance  of  the  two  windings  reduced  to  one  of  the  circuits 
that  is  of  importance.  Since  resistances  and  reactances  can  be 
transferred  from,  the  primary  circuit  to  the  secondary  or  vice 
versa,  when  multiplied  by  the  square  of  the  ratio  of  the  numbers 
of  turns  (Art.  446),  the  equivalent  inductance,  reduced  to  the 
primary  circuit,  and  per  leg  of  the  core,  is 


henrys.      .     (143) 


212  THE  MAGNETIC  CIRCUIT  [ART.  64 

All  the  lengths  here  are  expressed  in  centimeters,  and  /*=  1.257, 
also  a=di  +a2  is  the  spacing  between  the  coils,  which  is  a  known 
quantity.  Thus,  the  unknown  distances  ax  and  0%  which  enter 
into  the  expressions  for  LI  and  L2  are  eliminated  from  the 
formula  for  the  equivalent  inductance. 

The  coefficient  k  corrects  for  the  difference  between  the 
actual  linkages  shown  in  Fig.  50  at  the  left,  and  the  assumed 
linkages  shown  at  the  right.  The  values  of  k,  found  from  experi- 
ments, vary  within  quite  wide  limits,  depending  upon  the  pro- 
portions of  the  coils.  For  good  modern  transformers  Arnold 
gives  the  limits  of  k  between  0.95  and  1.05.1  See  also  eq.  (147) 
below.  Formula  (143)  gives  the  inductance  of  one  leg  only; 
the  equivalent  inductance  of  the  whole  transformer  depends 
upon  the  electrical  connections  between  the  coils. 

In  designing  a  transformer  the  coils  are  usually  arranged 
in  such  a  way  as  to  reduce  the  leakage  reactance  to  the  least 
possible  amount.2  Eq.  (143)  shows  that  in  order  to  achieve 
this  result,  a  comparatively  small  number  of  turns  must  be  used, 
and  the  coils  must  be  thin  and  long.  The  space  a  between  the 
coils  must  be  kept  as  small  as  is  compatible  with  the  require- 
ments for  insulation  and  cooling. 

The  usual  arrangement  of  coils  shown  in  Fig.  50  gives  a 
considerably  smaller  leakage  inductance  than  the  simpler  arrange- 
ment shown  in  Fig.  12.  Namely,  with  the  arrangement  shown 
in  Fig.  12,  the  permeance  of  the  path  of  the  complete  linkages 
in  the  space  between  the  coils  is  fj.aiOm/l.  This  expression 
differs  from  that  used  before  in  that  I  stands  in  the  denominator 
in  place  of  21.  For  the  partial  linkages  np  =  nl(x/b1),  where 
x  is  measured  now  from  the  edge  of  the  primary  coil,  furthest 
from  the  secondary  coil.  Thus,  the  primary  inductance  is  in 
this  case 


n       a, 

or 


By  symmetry  we  can  write  the  expression  for  L2,  and  hence, 

!E.  Arnold  Wechselstromtechnik  (2d  edition),  Vol.  1.,  p.  561. 
2  In  some  cases  a  considerable  leakage  reactance  is  specified  as  a  protection 
against  violent  short  circuits. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  213 

after  combining, 

.     .     .     (144) 


This  value  is  between  two  and  four  times  as  large  as  the 
value  given  by  eq.  (143).  For  this  reason,  in  most  transformers, 
the  low-tension  coil  is  split  into  two  sections;  compare  also  with 
Fig.  14. 

Formula  (143)  and  the  values  of  k  given  above  have  been 
deduced  for  the  core-type  transformer.  It  is  clear,  however, 
that  the  same  formulae  will  apply  to  the  shell-type  and  the  cruci- 
form type  transformers  with  cylindrical  coils,  though  the  coeffi- 
cient k  may  have  different  values  in  each  case.  Until  more 
reliable  and  numerous  experimental  data  are  available  the  same 
values  of  k  will  have  to  be  used  for  these  types  as  for  the  core- 
type.1 

(2)  Flat  Coils.  With  flat  coils  (Fig.  51)  the  inductance  of 
a  part  of  the  winding  between  AB  and  CD  can  be  calculated 
in  precisely  the  same  way  as  in  Fig.  50.  If  the  primary  winding 
is  split  into  q  sections,  the  inductance  per  section,  by  analogy 
with  eq.  (143),  is 

Leq/q=k[fjL(n1/q)2Om/2l][a  +  i(b1+b2)]lO-^henTys}   .    (145) 

where  the  dimension  I  is  again  measured  in  the  direction  of 
the  lines  of  force  and  Om  is  the  mean  length  of  a  turn.  The 
dimensions  a,  &x,  and  62  are  indicated  in  Fig.  51.  The  inductance 
of  the  whole  winding  is 

Leq=k(/jLnl2Om/2ql)[a+±(b1  +  b2)]W-*henrys,     .     .     (146) 

where  all  the  lengths  are  expressed  in  centimeters,  and  /*=  1.257. 

This  formula  presupposes  that  the  m.m.fs.  are  balanced,  or 
in  other  words,  that  there  are  two  half-sections  of  the  same 
winding  at  the  ends;  such  is  usually  the  case  in  order  to  reduce 
the  leakage  reactance.  (See  also  Fig.  13.) 

Eq.  (146)  shows  that  the  leakage  reactance  is  considerably 
reduced,  and  consequently  the  voltage  regulation  improved,  by 
subdividing  the  windings  and  placing  the  '  primary  and  the 

1  See  also  the  Standard  Handbook  for  Electrical  Engineers  under  Trans- 
former,  leakage  reactance. 


214 


THE  MAGNETIC   CIRCUIT 


[ART.  64 


secondary  windings  on  the  core  alternately.  At  a  given  voltage, 
and  with  a  given  type  of  construction,  the  spacing  a  between 
the  coils  may  be  considered  as  constant  and  independent  of  the 
number  of  sections.  In  transformers  for  extra-high  voltages  a 
is  large  as  compared  to  J(61+62),  so  that  the  leakage  reactance 
is  almost  inversely  proportional  to  the  number  of  sections  q. 
In  low-voltage  transformers  a  is  small  as  compared  to  61  and 
62;  hence,  Leq  is  almost  inversely  proportional  to  q2,  because 
bi  and  62  are  themselves  inversely  proportional  to  q.  Thus, 


FIG.  51. — The  leakage  field  in  a  transformer  with  flat  coils. 

in  general,   the  inductance  of  a  transformer  is  inversely  pro- 
portional to  qn,  where  n  has  a  value  between  1  and  2.1 

Dr.  W.  Rogowski  has  given  an  exact  mathematical  solution 
for  the  flux  distribution  in  the  case  of  flat  transformer  coils, 
assuming  the  coils  and  the  core  to  be  indefinitely  long  in  the 
direction  perpendicular  to  the  cross-section  shown  in  Fig.  5 1.2 

1  For  experimental   data  in  regard  to  the  effect  of  the  subdivision  and 
arrangement  of  transformer  windings  upon  the  leakage  reactance  see  Dr.  W. 
Rogowski,  Mitteilungen  Ueber  Forschungsarbeiten,  Heft  71   (Springer,  1909), 
p.  18,  also  his  article  Ueber  die  Streuung  des  Transformators,  Elektrotechnische 
Zeitschrift,  Vol.  31  (1910),  pp.  1035  and  1069;    also  Faccioli,   Reactance  of 
Shell-type  Transformers,   Electrical  World,  Vol.  55  (1910),  p.  941. 

2  Dr.  W.  Rogowski,  loc.   cit. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  215 

With   certain  simplifying  assumptions  he  arrived  at  the  same 
formula  for  inductance  as  eq.  (146)  in  which  approximately 


Because  of  the  simplifying  assumptions  made  in  the  deduc- 
tion of  this  formula,  the  values  of  k  calculated  from  the  results 
of  tests  on  actual  transformers  differ  slightly  from  those  given 
by  the  formula.  Let  k'  be  an  empirical  correction  coefficient, 
then 

]  .....     (147) 

In  Dr.  Rogowski's  experiments  the  actually  measured  induct- 
ance was  on  the  average  6  per  cent  higher  than  the  calculated 
one.  Until  more  experimental  data  are  available  it  is  therefore 
advisable  to  use  in  eq.  (147)  the  value  of  A;7  =1.06.  Eq.  (147) 
is  applicable  to  transformers  of  all  the  three  types  (Figs.  12  to 
14),  though  in  the  case  of  a  shell-type  or  cruciform  transformer, 
the  presence  of  iron  outside  the  coils  tends  to  increase  the  value 
of  &'.  However,  Dr.  Rogowski  states,  that  with  the  space  usually 
allowed  for  insulation  between  the  coils  and  the  iron,  the  influence 
of  the  iron  in  increasing  the  leakage  reactance  is  negligible.  Eq. 
(147)  holds  approximately  true  for  cylindrical  coils  also,  though 
there  are  asx  yet  no  conclusive  tests  for  the  value  of  the  cor- 
rection factor  to  be  used  with  such  coils. 

The  general  similarity  between  the  equations  for  leakage  induct- 
ance given  above  raises  the  question,  as  to  what  element  they 
possess  in  common.  This  is  found  in  the  conception  of  a  leakage 
coil,  which  is  the  "  fictitious  coil  "  spoken  of  above.  An  inspec- 
tion of  Figs.  50  and  51  will  show  that  the  successive  lines  of 
force  converge  upon  lines  which  may  be  called  the  "  hearts  " 
of  the  flux  system.  These  hearts  are  located  in  places  where 
the  net  m.m.f.  is  zero.  This  is  at  the  edge  of  the  half-coils  and 
at  the  center  of  the  whole-coils,  in  the  two  figures  mentioned. 
In  deriving  eq.  (144)  for  the  case  where  the  coils  are  not  split, 
the  heart  is  assumed  to  be  at  the  edge  of  both  coils.  If  we 
define  a  leakage  coil  as  that  part  of  the  winding  between  two 
successive  hearts,  then  eq.  (144)  will  always  apply  to  it.  In 
eq.  (143)  6X  and  62  refer  to  the  width  of  the  double  leakage  coil, 
hence  if  we  substitute  in  eq.  (144)  %bi  and  J62  for  &i  and  b2, 


216  THE  MAGNETIC  CIRCUIT  [ART.  64 

we  get  the  coefficient  J,  which  appears  in  eq.  (143).  In  eq. 
(143)  ni  and  Leg  refer  to  the  double  leakage  coil.  Substituting 
in  eq.  (144)  \n^  for  n±  and  %Leq  for  Leq,  we  get  eq.  (143).  Thus, 
the  differences  between  the  two  equations  is  readily  explained. 
In  case  the  coils  are  divided  in  any  unusual  manner,  we  must 
first  locate  the  hearts  by  noticing  where  the  m.m.fs.  are  balanced. 
Then  we  should  figure  out  the  inductance  by  eq.  (144)  for  each 
leakage  coil  separately.  The  only  precaution  to  be  observed  is 
that  the  various  quantities  refer  to  the  leakage  coil.  Finally 
(if  the  coils  are  in  series)  we  should  add  the  various  induct- 
ances together.  The  arrangement  with  half  coils  on  the  ends 
gives  the  minimum  of  inductance  for  a  given  number  of  coils. 

Prob.  1.  The  approximate  assumed  dimensions  of  a  1 5-kw.,  2200/1 1 0- 
v.,  60-cycle,  cruciform-type  transformer  with  cylindrical  coils  (Fig.  14) 
are:  Om  =140  cm.;  ^=4.5  cm.;  b2  =  3  cm.;  a  =  l  cm.  The  maximum 
useful  flux  is  1.03  megalines.  Show  that  the  relationship  between  the 
height  I  of  the  winding  and  the  percentage  reactive  voltage  drop  is 
a^  =  166.  Assume  fc  =  1.10. 

Prob.  2.  Referring  to  the  preceding  problem,  what  is  the  permeance 
of  the  space  between  the  outside  low-tension  coil  and  the  high-tension 
coil,  per  centimeter  of  the  height  of  the  coils,  and  what  is  the  effective 
value  of  the  flux  density  in  this  space,  at  full  load  ? 

Ans.     197.5  perms  per  cm.,  3420 /I  lines  per  sq.cm. 

Prob.  3.  Each  leg  of  a  core-type  transformer  is  provided  with  six 
flat  high-tension  coils  of  530  turns  each,  interposed  with  the  same 
number  of  low-tension  coils  of  40  turns  each,  one  of  the  low-tension 
coils  being  split  in  two  and  placed  at  the  ends.  The  high-tension  coils 
are  wound  of  3  mm.  round  wire,  53  layers,  10  turns  per  layer  (61  =  3  cm.) ; 
the  low-tension  coils  are  wound  of  8  mm.  square  wire,  in  20  layers,  2  turns 
per  layer  (62  =  1.6  cm.).  The  distance  between  the  coils  is  20  mm. 
Taking  the  inductance  of  this  transformer  to  be  unity,  calculate  the 
relative  inductances  of  the  transformer  when  the  high-tension  winding 
is  divided  into  three  coils  and  also  into  two  coils,  assuming  fc,  I,  and 
a  to  be  the  same  in  all  cases.  Ans.  2.55;  4.66. 

Prob.  4.  Solve  the  preceding  problem,  taking  into  account  the 
change  in  k.  Ans.  2.42;  4.14. 

Prob.  6.  The  following  results  were  obtained  from  a  short-circuit 
test  on  a  22/2-kv.,  2500-kva.,  60-cycle,  shell-type  transformer,  with 
flat  coils:  With  the  high-tension  winding  short-circuited,  and  full  rated 
current  flowing  through  the  low-tension  winding  the  voltage  across 
the  secondary  terminals  was  73.5  v.,  and  the  wattmeter  reading  was 
27  kw.  The  transformer  winding  consists  of  12  high-tension  coils  of 
100  turns  each,  and  of  11  low-tension  coils  interposed  between  the 
high-tension  coils,  together  with  2  half-coils  at  the  ends.  The  dimen- 
sions of  the  coils  are :  Om=2.6m.;  Z  =  18cm.;  61  =  16mm.;  62  =  1 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  217 

a  =  12  mm.    Calculate  the  correction  factor  k'  in  formula  (147).    Hint: 
Eliminate  the  ir  drop,  using  the  wattmeter  reading.    Ans.    About  1.05. 

Prob.  6.  What  is  the  greatest  permissible  thickness  of  the  coils 
of  a  60-cycle  transformer,  if  the  reactive  drop  must  not  be  larger  than 
three  times  the  resistance  drop?  The  ducts  a  are  1  cm.  The  space 
factor  of  the  copper  in  each  coil  is  0.55.  The  primary  coils  and  the 
whole  coils  of  the  secondary  are  of  the  same  thickness.  fc  =  0.98. 
Hint :  Assume  Om,  I,  and  nt  and  show  that  they  cancel  out. 

Ans.     b  =  3.66  cm. 

Prob.  7.  In  order  to  provide  a  better  cooling,  and  at  the  same 
time  save  on  insulation,  two  flat  high-tension  coils  are  frequently  placed 
side  by  side,  with  a  small  air-space  in  between,  and  in  the  same  way 
the  low-tension  coils  may  be  subdivided.  Show  that  no  leakage  flux 
passes  in  the  space  between  the  two  adjacent  coils  which  belong  to 
the  same  winding,  so  that  the  inductance  of  the  winding  is  not  appreciably 
increased  by  these  spaces,  and  can  be  calculated  as  if  these  spaces  did 
not  exist.1 

Prob.  8.  Compare  the  formulae  given  above  and  the  numerical  values 
of  transformer  leakage  reactance  obtained  therefrom  with  the  formulae 
and  data  given  in  the  Standard  Handbook  for  Electrical  Engineers.  Do 
this  for  any  transformer,  the  dimensions  of  which  are  available. 

Prob.  9.  The  equivalent  reactance  of  a  transformer  is  Zj  reduced 
to  the  primary  circuit,  and  is  x2  reduced  to  the  secondary  circuit.  All 
the  primary  coils  are  connected  in  series,  and  all  the  secondary  coils 
are  also  in  series.  Show  that  these  equivalent  reactances  become 
XI/GI  and  x2/c22  respectively,  when  the  primary  winding  is  divided 
into  cx  branches  in  parallel,  and  the  secondary  winding  is  divided  into 
c2  branches  in  parallel.  The  division  is  supposed  to  be  made  in  such 
a  way  as  to  keep  the  m.m.fs.  in  the  adjacent  coils  balanced.  Hint: 
If  the  equivalent  reactance  of  one  primary  branch  is  z/,  that  of  c-^ 
branches  in  series  is  x1  =  x1c/,  and  that  of  cx  branches  in  parallel  is  a^'/Ci 
no  matter  whether  the  secondary  coils  are  connected  in  series  or  in 
parallel. 

Prob.  10.  Prove  that  the  equivalent  reactance  of  a  transformer 
is  the  same,  whether  the  coils  are  in  series  or  in  parallel,  provided  that 
the  total  number  of  turns  in  series  is  the  same. 

Note:  Sometimes,  because  of  the  difficulty  in  using  heavy  con- 
ductors, it  is  desirable  to  multiple  the  coils.  In  such  case,  the  parallel 
coils  must  have  the  same  number  of  turns,  and  they  should  be  sym- 
metrically arranged,  so  as  to  prevent  exchange  currents. 

65.  The  Equivalent  Leakage  Permeance  of  Armature  Windings. 
In  certain  problems  relating  to  the  design  and  operation  of 
electrical  machinery  it  is  necessary  to  calculate  the  inductance 

1  This  fact  has  been  verified  experimentally;  see  Arnold,  Wechselstrom- 
technik,  Vol.  2  (1910),  p.  29. 


218 


THE  MAGNETIC  CIRCUIT 


[ART.  65 


of  the  armature  windings.  This  inductance  affects  the  per- 
formance of  the  machine,  because  the  leakage  fluxes  created 
by  the  armature  currents  induce  e.m.fs.  in  the  machine.  Such 
leakage  fluxes  are  shown  in  Figs.  23  and  36,  in  an  induction 
machine  and  a  synchronous  machine  respectively.  For  purposes 
of  theory  and  computation  these  leakage  fluxes  are  usually 
subdivided  into  three  parts,  namely: 

(a)  Leakage  fluxes  linked  with  the  parts  of  the  windings 
embedded  in  the  armature  iron  (Figs.  36  and  54).     These  paths 
are  closed  partly  through  the  slots,  and  partly  through  the  tooth- 
tips  (slot  leakage  and  tooth-tip  leakage). 

(b)  Leakage  fluxes  linked  with  the  parts  of  the  armature 
windings  in  the  air-ducts. 


> 

•N 

CEP 
S 

>  N 

'S 

FIG.  52. — Undivided  end-connections.        FIG.  53. — Divided  end-connections. 


(c)  Leakage  fluxes  linked  with  the  end-connections  of  the 
armature  windings  (Figs.  52  and  53). 

Usually  the  fluxes  (a)  and  (b)  are  merely  distortional  com- 
ponents of  the  main  flux  of  the  machine,  and  only  the  fluxes 
(c)  have  a  real  existence. 

It  will  be  readily  seen  that  the  paths  of  the  tooth-tip  leakage 
and  of  that  around  the  end-connections  are  too  complicated 
to  allow  the  corresponding  permeances  to  be  calculated  theo- 
retically. For  this  reason,  various  empirical  and  semi-empirical 
formula  are  used  in  practice  for  estimating  the  leakage  inductance 
of  armature  windings,  the  coefficients  in  these  formulae  being 
determined  from  tests  on  similar  machines. 

The  most  rational  procedure  is  to  express  the  inductance 
through  the  equivalent  permeance  of  the  paths,  as  defined  by 
eq.  (106a)  in  Art.  58.  Let  there  be  CPP  conductors  per  pole 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  219 

per  phase,  such  as  are  indicated  for  instance  in  Fig.  15.  Then 
the  inductance  of  a  machine,  per  pole  per  phase,  is  given  by  the 
equation 

LPP=CPP2(Peg,      .    .    ....     (148) 

where  (Peq  is  an  empirical  value  of  the  equivalent  permeance 
per  pole  per  phase.  This  formula  presupposes  that  all  the 
partial  linkages  are  replaced  by  the  equivalent  complete  linkages 
embracing  all  the  Cpp  conductors.  Moreover,  the  value  of  (Peq 
is  such  as  to  take  into  account  the  inductive  action  of  the  other 
phases  upon  the  phase  under  consideration.  The  total  inductance 
of  the  machine,  per  phase,  depends  upon  the  electrical  connec- 
tions in  the  armature  winding.  If  all  the  p  poles  are  connected 
in  series,  the  foregoing  expression  for  LPP  must  be  multiplied 
by  p;  if  there  are  two  branches  in  parallel,  the  inductance 
of  each  branch  is  %pLPP,  and  the  combined  inductance  of  the 
whole  machine  is  %(%pLpp)  =  \pLPP. 

The  leakage  inductance  of  a  machine  is  usually  determined 
by  sending  through  it  an  alternating  current  of  a  known  frequency, 
under  conditions  which  depend  upon  the  kind  of  the  machine 
(the  field  to  be  removed  in  a  synchronous  machine,  and  the 
armature  to  be  locked  in  an  induction  machine).  From  the 
readings  of  the  current  of  the  applied  voltage  and  the  watts 
input,  the  reactance  x  of  the  machine  is  calculated  (after  elim- 
inating the  ohmic  drop) .  Then,  knowing  the  frequency  /  of  the 
supply  and  the  number  of  poles  of  the  machine,  the  inductance 
LPP=x/(2xfp)  per  pole  is  calculated.  Substituting  into  eq. 
(148)  this  value  of  Lpp  and  the  known  number  of  conductors 
Cpp,  the  equivalent  permeance  (Peq  per  pole  per  phase  is  deter- 
mined. By  performing  such  tests  on  machines  built  on  the 
same  punching,  but  of  different  embedded  lengths,  the  permeance 
due  to  the  embedded  parts  of  the  winding  is  separated  from 
that  due  to  the  end-connections;  the  values  so  obtained  are  then 
used  in  new  designs. 

The  leakage  permeances  in  the  embedded  parts  are  pro- 
portional to  the  widths  of  these  parts  in  the  direction  parallel 
to  the  shaft,  in  other  words,  to  the  length  of  conductors  which 
are  surrounded  by  the  leakage  lines.  Experiment  shows  that 
the  permeance  of  the  paths  in  the  air-ducts  and  around  the 
end-connections  is  also  approximately  proportional  to  the  lengths 


220  THE   MAGNETIC   CIRCUIT  [ART.  65 

of  the  corresponding  parts  of  the  armature  coils.     Since  all  these 
permeances  are  in  parallel,  (Peq  is  equal  to  their  sum,  or 

e          ....       (149) 


Here  the  letter  I  denotes  the  lengths  in  cm.  of  the  coil,  or,  what 
is  the  same  thing,  the  width  of  the  paths  of  flux.  The  subscripts 
i,  a,  and  e  refer  to  the  iron,  the  air-ducts,  and  the  end-connections 
respectively.  Thus,  ^  is  the  semi-net  length  of  the  core,  that  is, 
the  length  exclusive  of  ducts  but  inclusive  of  the  space  between 
the  laminations.1  The  corresponding  permeance  per  centimeter  is 
(Pj.  The  sign  "  prime  "  signifies  that  the  corresponding  <P's  refer 
to  one  centimeter  width  of  path. 

The  coefficient  a.  is  equal  to  1  when  the  end-connections 
are  arranged  as  in  Fig.  52,  and  a=J  when  they  are  arranged 
according  to  Fig.  53.  Namely,  in  the  first  case  the  number  of 
conductors  Ce  per  group  of  the  end-connections  is  equal  to  the 
number  CPP  in  the  embedded  part.  In  the  second  case  Ce  is 
equal  to  JCPP.  In  the  first  case  there  are  as  many  groups  of 
end-connections  per  phase  as  there  are  poles;  in  the  second 
case  there  are  twice  as  many  groups  per  phase  as  there  are  poles, 
so  that  two  groups  (one  pointing  to  the  right  and  one  to  the  left) 
must  be  counted  per  pole.  Thus,  with  undivided  end-con- 
nections, the  inductance  is  Cpp2(pe'le,  while  with  divided  end- 
connections  it  is  (%CPP)2(Pef.2le=Cpp2.  \(Pele-  This  accounts  for 
the  value  of  a  =  J,  in  formula  (149),  and  shows  that  the  inductance 
due  to  the  end-connections  is  reduced  twice  by  subdividing 
them  into  two  groups.  When  estimating  the  leakage  reactance 
it  is  therefore  necessary  to  know  the  exact  arrangement  of  the 
end-connections. 

In  preliminary  calculations,  before  the  armature  coils  are 
drawn  to  scale,  the  length  le  of  the  end-connections  in  a  full- 
pitch  winding  is  usually  assumed  to  be  equal  to  about  1.5r,  where 
T  is  the  pole  pitch.  For  fractional-pitch  windings,  le  varies 
roughly  as  the  winding-pitch,  or  Ze=1.5^r  (see  Art.  29). 

1  The  semi-net  length  is  used  in  getting  the  leakage  permeance  in  the 
slot,  because  the  flux  spreads  as  it  comes  out  of  the  iron  almost  immediately. 
The  spaces  between  the  laminations  do  not  affect  the  density  in  the  air 
because  they  are  so  small  as  compared  to  the  dimensions  of  the  slot. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  221 

Substituting  the  value  of  (Peq  from  eq.  (149)  into  eq.  (148) 
we  obtain 

Lpp=CPp2((Pifli+(Pafla  +  ^efle^~8  he*ryS.         .        (150) 

In  the  following  three  articles  formula  (150)  is  applied  to 
the  calculation  of  the  leakage  reactance  of 

(a)  Induction  machines; 

(6)   Synchronous  machines; 

(c)    Coils  undergoing  commutation  in  a  direct-current  machine. 

In  each  case  somewhat  different  values  of  the  unit  permeances 
(Pf  are  used,  because  of  the  diversity  of  the  magnetic  paths. 

66.  The  Leakage  Reactance   in  Induction  Machines.     It  is 

explained  in  Art.  35  and  shown  in  Fig.  23  that  the  actual  flux 
in  a  loaded  induction  machine  is  the  resultant  of  three  fluxes, 
of  which  the  useful  flux  $  is  linked  with  both  the  primary  and 
the  secondary  windings.  The  component  fluxes  4>l  and  $2, 
linked  with  the  primary  and  secondary  windings  respectively, 
are  called  the  leakage  fluxes.  They  induce  in  the  windings  e.m.fs. 
in  quadrature  with  the  corresponding  currents,  and  these  e.m.fs. 
have  to  be  balanced  by  a  part  of  the  terminal  voltage.  Con- 
sequently, that  part  of  the  applied  voltage  which  is  balanced  by 
the  useful  flux  is  reduced;  in  other  words,  the  useful  flux  and 
the  useful  torque  are  reduced  with  a  given  current  input.  As 
a  matter  of  fact,  the  maximum  torque  and  the  overload  capacity 
of  an  induction  machine  are  essentially  determined  by  its  leakage 
fluxes,  or  what  amounts  to  the  same  thing,  its  leakage  inductances. 

Knowing  the  primary  and  secondary  leakage  reactances,  the 
actual  induction  machine  is  replaced  by  an  equivalent  electric 
circuit  (or  a  circle  diagram  is  drawn  for  it),  after  which  its 
performance  can  be  predicted  at  any  desired  load.  The  problem 
here  is  to  determine  the  values  of  these  leakage  reactances  and 
inductances,  from  the  given  dimensions  of  a  machine.  The 
rest  of  the  problem  is  treated  in  the  Electric  Circuit. 

The  leakages  fluxes,  which  are  indicated  schematically  in 
Fig.  23,  are  shown  more  in  detail  in  Fig.  54.  The  primary 
conductors  in  one  of  the  phases  and  under  one  pole  are  marked 
with  dots,  and  the  adjacent  secondary  conductors  are  marked 
with  crosses,  to  indicate  the  currents  which  are  flowing  in  them. 

Assuming   the   rotor   to   be    provided    with    a    squirrel-cage 


222 


THE  MAGNETIC  CIRCUIT 


[ART.  66 


winding,  the  distribution  of  the  secondary  currents  is  prac- 
tically an  image  of  the  primary  currents.  Neglecting  the  mag- 
netizing ampere-turns  necessary  for  establishing  the  main  or 
useful  flux,  the  secondary  ampere-turns  per  pole  per  phase  are 
equal  and  opposite  to  the  primary  ampere-turns.  A  similar 
assumption  is  also  made  in  the  preceding  article,  in  the  case  of 
the  transformer.  This  assumption  is  not  as  accurate  in  the 
case  of  an  induction  machine,  because  here  the  magnetizing 
current  is  proportionately  much  larger,  due  to  the  air-gap; 
nevertheless,  the  assumption  is  sufficiently  accurate  for  most 


FIG.  54. — The  slot  and  zig-zag  leakage  fluxes  in  an  induction  machine. 

practical  purposes.  Even  if  the  magnetizing  current  is  equal 
to  say  25  per  cent  of  the  full-load  current,  the  difference  between 
the  primary  and  the  secondary  ampere-turns  should  be  less 
than  10  per  cent,  because  the  magnetizing  current  is  considerably 
out  of  phase  with  the  secondary  current.1 

With  this  assumption,  the  primary  and  the  secondary  current 
belts  shown  in  Fig.  54  may  be  considered  as  two  sides  of  a  narrow 
fictitious  coil  which  excites  the  leakage  flux,  causing  it  to  pass 
circumferentially  along  the  active  layer.2  Neglecting  the  mutual 

1  See  the  circle  diagram  of  an  induction  motor,  for  instance,  in  the  author's 
Experimental  Electrical  Engineering,  Vol.  2,  p.  167. 

2  Although  the  secondary  frequency  is  different  from  the  primary,  with 
respect  to  the  revolving  rotor  it  is  the  same  as  the  primary  frequency  with 
respect  to  the  stator.     Let  s  be  the  slip  expressed  as  a  fraction  of  the  primary 
frequency.     Then  the  speed  of  the  rotor  is  (1  —  s),  and  the  frequency  of  the 
secondary  currents  with  respect  to  a  fixed  point  on  the  stator  is  s  +  (1  —  s)  =  1. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  223 

action  of  the  consecutive  phases,  the  length  of  this  flux  is  approx- 
imately r/m,  where  r  is  the  pole  pitch  and  m  is  the  number 
of  the  stator  phases.  Part  of  the  flux  is  linked  with  the  primary 
current  belt,  and  part  with  the  secondary  belt.  The  conditions 
are  essentially  the  same  as  between  the  transformer  windings 
P1  and  $!  in  Fig.  50.  Knowing  the  equivalent  permeances 
of  the  individual  paths  the  inductance  can  be  calculated  from 
eq.  (150). 

In  an  induction  machine  with  a  squirrel-cage  rotor  the  total 
leakage  in  the  embedded  part  may  be  resolved  into  three  com- 
ponents shown  in  Fig.  54,  namely : 

(1)  The  primary  slot  leakage,  4>8l; 

(2)  The  secondary  slot  leakage,  tf>s2; 

(3)  The  tooth-tip  or  zigzag  leakage,  ®z. 

The  fluxes  tf>sl  and  4>s2  are  alternating  fluxes  of  the  frequency 
of  the  corresponding  currents.  The  zigzag  flux  ®z  varies  according 
to  a  much  more  complicated  law,  because  the  permeance  of  its 
path  changes  from  instant  to  instant  in  accordance  with  the 
relative  position  of  the  stator  and  rotor  teeth;  compare  posi- 
tions (1)  and  (2)  in  Fig.  23.  Moreover,  Fig.  54  shows  only  the 
simplest  case,  which  never  occurs  in  practice,  namely;  when 
the  stator  tooth  pitch  is  equal  to  that  in  the  rotor.  In  reality, 
the  two  pitches  are  always  selected  so  as  to  be  different,  in  order 
to  avoid  the  motor  sticking  at  sub-synchronous  speeds  (due  to  the 
higher  harmonics  in  the  fluxes  and  in  the  currents).  Therefore, 
the  paths  of  the  zigzag  leakage  flux  are  much  more  complicated 
than  is  shown  in  Fig.  54,  and  in  calculations  the  average  per- 
meance of  the  zigzag  path  is  used. 

In  a  machine  with  a  phase-wound  rotor  the  main  flux  is 
further  distorted,  due  to  the  fact  that  the  primary  and  secondary 
phase-belts  are  not  exactly  in  space  opposition  at  all  moments. 
While  the  total  m.m.fs.  of  the  primary  and  secondary  are 
balanced,  there  is  a  local  unbalancing  which  changes  from 
instant  to  instant.  This  distortion  is  the  same  as  if  it  were  due 
to  an  additional  leakage,  which  was  named  by  Professor  C.  A. 
Adams  the  belt  leakage.1  This  part  of  the  leakage  usually 
constitutes  but  a  small  part  of  the  total  leakage,  and  will  not 
be  considered  here  separately.  Those  interested  are  referred  to 

1  C.  A.  Adams,  The  Leakage  Reactance  of  Induction  Motors,  Trans. 
Intern.  Electr.  Congress,  St.  Louis,  1904,  Vol.  1,  p.  711. 


224  THE  MAGNETIC  CIRCUIT  [ART.  66 

the  original  paper  and  to  the  works  mentioned  at  the  end  of 
this  article. 

Let  there  be  CPP1  conductors  per  pole  per  phase  in  the  stator 
winding;  then  the  fictitious  coil  (Fig.  54)  made  up  of  the 
primary  and  secondary  conductors  has  CPP1  turns,  when  reduced 
to  the  primary  circuit.  This  is  because  the  secondary  winding 
can  be  replaced  by  an  equivalent  winding  with  a  "  one  to  one  " 
ratio,  that  is,  with  the  same  number  of  conductors  as  the 
primary  winding.  In  this  case,  the  secondary  current  is  equal 
to  the  primary  current  (Art.  446).  Therefore,  eq.  (150)  can 
be  made  to  give  the  equivalent  inductance,  including  the  pri- 
mary inductance  and  the  secondary  inductance  reduced  to  the 
primary  circuit,  provided  that  the  permeances  of  the  paths 
linking  with  the  secondary  conductors  are  included  in  the  values 
of  (P"s.  Such  is  naturally  the  case  when  the  values  are  deter- 
mined from  a  test  with  the  rotor  locked. 

Extended  tests  have  shown  that  in  a  given  line  of  machines 
the  permeance  (Pj  is  inversely  proportional  to  the  peripheral 
length  of  the  equivalent  leakage  flux,  that  is,  inversely  proportional 
to  (r/w),  where  T  is  the  pole  pitch  and  ra  is  the  number  of 
primary  phases.1  This  shows  that  the  permeance  per  centi- 
meter of  peripheral  length  of  the  active  layer  is  fairly  constant, 
in  spite  of  different  dimensions  and  proportions,  as  long  as 
these  are  varied  within  reasonable  limits.  Thus 


where  G>"  is  the  leakage  permeance  of  the  active  layer  in  the 
embedded  part  per  one  centimeter  of  axial  length  and  per  centi- 
meter of  the  peripheral  length  of  the  path.  Thus,  the  final 
formula  for  the  equivalent  leakage  inductance  of  an  induction 
machine,  per  pole  per  phase,  reduced  to  the  primary  circuit,  is 

Z^PP=  CPP12[CP/V(V™)  +  (Pa'la  +  a(Pe'le]W-*  henrys.    (151) 

In  this  formula  the  following  average  values  of  unit  per- 
meance may  be  used  for  machines  of  usual  proportions,  unless 
more  accurate  data  are  available. 

1  H.  M.  Hobart,  Electric  Motors  (1910),  table  on  p.  397.  The  values 
for  (P/'  given  below  have  been  computed  from  this  table,  and  the  results 
multiplied  by  2,  because  the  table  gives  the  values  of  the  primary  per- 
meances only. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  225 

The  equivalent  permeance  (Pi"  of  the  embedded  part  in  perms 
per  centimeter  of  the  semi-net  axial  length  of  the  machine,  and 
per  centimeter  of  peripheral  length  of  the  air  gap,  is  for 

Open  slots  Half-open  slots     Completely  closed  slots 

11.5  14.5  18. 

The  equivalent  permeance  around  the  end-connections,  and 
around  the  parts  of  the  conductors  in  the  air-ducts,  decreases 
with  the  increasing  number  of  slots  per  pole  per  phase,  for  the 
same  reason  that  the  slot  permeance  decreased.  In  induction 
motors  usually  at  least  three  slots  are  used  per  pole  per  phase, 
and  under  these  conditions  Mr.  H.  M.  Hobart  uses  <P/  =  0.8 
perm  per  centimeter,  with  phase-wound  rotors,  and  <P/  =  0.6 
perm  per  centimeter  with  squirrel-cage  rotors.  (Pa'  may  be 
taken  in  all  cases  equal  to  0.8  perm  per  cm.  The  lengths  le 
and  la  are  always  understood  to  refer  to  the  stator  winding. 

The  foregoing  data  refer  to  machines  with  full-pitch  windings 
in  the  stator  and  in  the  rotor.  With  a  fractional  pitch  winding 
the  equivalent  leakage  permeances  are  somewhat  smaller,  due  to 
longer  phase  belts  and  to  the  mutual  induction  of  the  over- 
lapping phases.  Let  the  winding-pitch  factor  (Art.  29)  of  the 
primary  winding  be  kwl  and  that  of  the  secondary  winding 
kw2.  Then  the  leakage  inductance  of  the  machine,  calculated 
for  a  full-pitch  winding  (but  for  Ze=1.5£r),  is  multiplied  by 
kwi-kW2-  This  is  an  empirical  correction,  which  is  accurate 
enough  for  ordinary  practical  purposes.  In  reality,  of  two 
machines  designed  for  a  given  duty,  one  with  a  fractional  pitch 
winding  and  the  other  not  (but  otherwise  both  alike)  the  first 
often  has  a  higher  inductance  than  the  second.  This  is  because 
more  turns  are  required  with  the  fractional  pitch  winding,  if  the 
flux  densities  in  the  iron  and  in  the  air-gap  are  to  be  the  same 
in  both  cases.1 

With  the  data  given  above  the  calculation  of  the  leakage 
reactance  of  a  given  induction  motor  is  quite  simple,  and  one 
engaged  regularly  in  the  commercial  design  of  induction-motors 
can  obtain  sufficiently  accurate  data  for  their  design  or  for  the 

1  For  a  theoretical  and  experimental  investigation  of  the  effect  of  a 
fractional  pitch  upon  the  leakage  reactance  in  induction  machines  see 
C.  A.  Adams,  Fractional-pitch  Windings  for  Induction  Motors,  Trans.  Amer. 
Inst.  Elec.  Engrs.,  Vol.  26  (1907),  p.  1488. 


226  THE   MAGNETIC  CIRCUIT  [ART.  66 

computation  of  their  performance,  provided  that  he  adapts  the 
numerical  values  of  the  unit  permeances  to  each  individual  case, 
on  the  basis  of  his  previous  experience.  Many  authors  have 
given  theoretical  formulae  for  the  leakage  inductance  of  induction 
machines.1  These  formulae,  curves,  and  methods,  while  useful 
in  accurate  work,  are  too  elaborate  to  be  quoted  here ;  at  any  rate 
they  are  of  interest  only  to  a  specialist  in  design.  Two  examples 
of  the  theoretical  calculation  of  leakage  inductance  are  given 
below,  in  order  to  show  the  student  the  general  method  used, 
and  thus  introduce  him  into  the  literature  on  the  subject. 

(a)  A  Theoretical  Calculation  of  the  Slot  Leakage  Permeance. 
We  shall  calculate  the  equivalent  permeance  for  the  half  closed 
slot  (Fig.  55),  an  open  slot  being  a  special  case  of  it.  With 
the  notation  shown  in  sketch,  and  with  SPP  slots  per  pole  per 
phase,  we  have: 

PC'  =  Ab2/s2  +  63/^3  +  bt'/8t]/SPP;    d(?  P'  =  fidx/  (s4SPP), 
and  nx=CPPx/b4.     Hence 


and 

.   .     (152) 


The  equivalent  permeance  of  one  slot,  per  unit  of  the  semi- 
net  axial  length  of  the  machine  is 

^.'=M&2/*2  +  &3/«3+(J&4  +  &40/*4].        -       -       -       (153) 

This   shows  that  the   slot   permeance   depends   only  upon    the 
proportions  of  the  slot  and  not  upon  its  absolute  dimensions. 

.(b)  A  Theoretical  Calculation  of  the  Zigzag  Leakage  Permeance. 
The  calculation  of  the  zigzag  leakage  permeance  is  simple  only 

1  C.  A.  Adams,  loc.  cit.;  also  Trans.  Amer.  Inst.  Elec.  Engrs.,  Vol.  24 
(1905),  p.  338;  ibid.,  Vol.  26  (1907),  p.  1488.  Hobart,  Electric  Motors, 
(1910),  Chap,  xxi;  Arnold,  Wechselstromtechnik,  Vol.  5,  part  1  (1909),  pp. 
49-54;  R.  Goldschmidt,  Appendix  to  his  book  on  The  Alternating  Current 
Commutator  Motor  (1909);  R.  E.  Hellmund,  Practische  Berechnung  des 
Streuungskoeffizienten  in  Induktionsmotoren,  \Elektrotechnische  Zeitschrift, 
Vol.  31  (1910),  p.  1111  and  1140;  W.  Rogowski,  Zur  Streuung  des  Dreh- 
strommotors,  ibid.,  pp.  356,  1292,  and  1316.  See  also  an  extensive  series 
of  articles  by  J.  Rezelman  in  La  Lumiere  Electrique,  1909-1911,  and  in 
the  (London)  Electrician. 


CHAP.  XII] 


INDUCTANCE  OF  WINDINGS 


227 


when  the  stator  tooth-pitch  and  the  rotor  tooth-pitch  are  equal 
to  each  other;  otherwise  the  paths  become  too  complicated  for 
mathematical  analysis.  Since  the  position  of  the  secondary  teeth 
varies  with  respect  to  the  primary  teeth,  the  permeance  of  the 
zigzag  leakage  also  varies,  and  it  is  necessary  to  take  its  average 
value  over  one-half  of  the  tooth-pitch  X,  that  is,  between  the  posi- 
tions (1)  and  (2)  in  Fig.  23.  In  some  intermediate  position  (Fig. 
55),  determined  by  the  overlap  y,  the  reluctance  of  the  path  /, 
per  unit  of  semi-net  axial  length  of  the  iron,  is  a/  (//?/)  rels. 


FIG.  55. — The  notation  used  in  the  calculation  of  the  slot  and  zigzag  leakage 

The  reluctance  of  the  path  g  is  a/fi(t2  —s1  —y)  rels.  The  com- 
bined reluctance  of  /  and  g  is  equal  to  the  sum  of  the  foregoing 
expressions.  The  permeance  of  /  and  g  in  series,  being  the 
reciprocal  of  the  combined  reluctance,  is 


The  permeance  in  question  varies  according  to  this  law,  for  the 
positions  of  the  secondary  tooth  between  ?/  =  i02~si)  and  2/  =  0, 
the  tooth  moving  to  the  left.  From  y=0  to*  2/=i(^2~si~~"^)  the 
permeance  is  practically  equal  to  zero,  because  the  secondary  tooth 
bridges  over  the  primary  slot  no  more.  The  student  is  advised 


228 


THE  MAGNETIC  CIRCUIT 


[ART.  66 


to  mark  the  positions  of  the  rotor  teeth  on  a  strip  of  paper  and 
to  place  them  in  different  positions  with  respect  to  the  primary 
slot  in  order  to  see  the  variations  in  the  overlap.  Thus,  the 
average  value  of  the  zigzag  permeance  per  tooth  pitch  is 


i(fc- 


(Pv'dy=p(t2-81)2/(6aX)  perms/cm.    (154) 


Prob.  11.  Draw  a  sketch  similar  to  Fig.  54,  but  with  the  rotor 
tooth-pitch  different  from  that  in  the  stator,  and  indicate  roughly  the 
general  character  of  the  paths  of  the  zigzag  leakage. 

Prob.  12.  Show  that  increasing  the  number  of  slots  per  pole  in  a 
given  machine  does  not  alter  materially  the  slot  leakage,  but  reduces 
considerably  the  zigzag  leakage. 

Prob.  13.  Calculate  the  equivalent  leakage  inductance  per  phase 
of  a  three-phase,  10-pole,  induction  machine,  with  15  slots  per  pole  in 
the  stator,  and  a  phase-wound  rotor.  Both  windings  have  100  per 
cent  pitch,  the  slots  are  semi-closed  on  both  punchings,  the  individual 
stator  coils  in  each  phase  are  connected  in  series,  and  there  are  20  con- 
ductors in  each  stator  slot.  The  bore  of  the  machine  is  110  cm.,  the  gross 
length  of  the  core  is  30  cm. ;  there  are  three  vents  of  8  mm.  each.  The 
end-connections  are  arranged  according  to  Fig.  53.  Ans.  57.5  mh. 

Prob.  14.  The  design  of  the  machine  in  the  preceding  problem 
has  been  modified  in  the  following  respects:  The  rotor  is  provided 
with  a  squirrel-cage  winding,  the  winding  pitch  in  the  stator  is  shortened 
to  80  per  cent,  the  stator  slots  are  made  open,  and  the  end  connections 
are  divided,  as  in  Fig.  54.  What  is  the  inductance  of  the  machine? 

Ans.    43.3  mh. 

Prob.  15.  Check  the  values  of  <P/'  given  in  the  text  above  with 
those  in  Hobart's  table. 

Prob.  16.    For  the  usual  limits  of  proportions  of  slots,  teeth,  and 

air-gap  calculate  the  values  of  (P/' 
from  eqs.  (153)  and  (154)  and  com- 
pare the  results  with  the  average 
experimental  values  given  in  this 
text. 

Prob.  17.      Calculate  the  equi- 
valent leakage  permeance  of  a  round 
"N      slot    (Fig.    56),    assuming  the  con- 
ductors   to    completely    fill  it,  and 
{  }      the  lines  of  force  to  be  straight  lines. 

^ •'       Hint:  Select  the  angle  a.  as  the  inde- 

FIG.  56. — A  semi-closed  round  slot,      pendent  variable,  and  integrate  eq. 

(106)  between     a  =  0    and    a  =  it. 
See  Arnold,  Wechselstromtechnik,  Vol.  4  (1904),  p.  44. 

Ans.    (5Y  =  //(0.623  +b/s)  perms  per  cm. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  229 

67.  The  Leakage  Reactance  in  Synchronous  Machines.     The 

physical  nature  of  the  armature  reactance  in  a  synchronous 
machine  is  explained  in  Art.  46;  the  influence  of  this  reactance 
upon  the  performance  of  a  machine  is  shown  in  Figs.  37,  38,  40, 
and  41.  The  problem  here  is  to  calculate  the  numerical  value 
of  this  reactance  for  a  given  machine,  using  eq.  (150)  with 
empirical  coefficients  (P'.1 

It  may  also  be  stated  here  that  for  standard  machines,  partic- 
ularly in  preliminary  estimates,  the  ix  drop  is  sometimes  taken 
as  a  certain  percentage  of  the  rated  voltage  of  the  machine 
instead  of  estimating  the  inductance  from  formula  (150).  In 
synchronous  generators  the  ix  drop  at  the  rated  volt-ampere 
load  varies  from  5  to  10  per  cent  of  the  rated  terminal  voltage. 
In  synchronous  motors,  where  some  inductance  is  useful,  the 
ix  drop  ranges  from  8  to  15  per  cent  of  the  rated  voltage.  For 
60-cycle  machines,  and  for  machines  with  a  comparatively  large 
number  of  armature  ampere-turns,  values  must  be  taken  nearer 
the  higher  limit.  For  25-cycle  machines,  and  for  machines 
with  a  comparatively  small  number  of  armature  ampere-turns, 
values  must  be  taken  nearer  the  lower  limit.  A  considerable 
error  in  estimating  the  value  of  ix  has  but  little  effect  upon  the 
calculated  performance  at  unity  power  factor,  because  the  vector 
ix  is  then  perpendicular  to  e  (Figs.  37,  38,  40  and  41).  However,  a 
considerable  error  may  be  introduced  at  lower  values  of  the 
power  factor  if  the  reactive  drop  ix  has  not  been  estimated 
with  a  sufficient  accuracy. 

The  values  of  (P'  for  synchronous  machines  are  different 
from  those  given  above  for  induction  machines,  because  of  the 
absence  of  any  secondary  current-belts.  Parshall  and  Hobart  2 
give  the  following  values  for  (P{ : 

1  For  a  theoretical  calculation  of  the  coefficient  (Pf,  see  Arnold,  Wechsel- 
stromtechnik,  Vol.  4  (1904),  pp.  41-52;    Hawkins  and  Wallis,    The  Dynamo, 
Vol.  2  (1909),  pp.  901-904.    For  a  comparison  between  the  calculated  and 
actually  measured  values  see  an  extended  series  of  articles  by  J.  Rezelman 
in  La  Lumiere  Electrique,  1909-1911,  and  in  The  (London)  Electrician. 

2  Electric  Machine  Design   (1906),  p.  478.     These  values  are  corroborated 
by  those  obtained  by  Pichelmayer;   see  his  Dynamobau,    1908,  pp.  208  and 
504.     Pichelmayer's  values  for  (Pj   (which  he  denotes  by   £)   are  somewhat 
high    because   the   end-connection    leakage   is    not   considered    separately. 
Arnold's  values,  given  in  his  Wechselstromtechnik,   Vol.  4,  p.  280,  should  be 
used  with  discretion,    because  they  apply  to  a  different  formula;    namely, 


230  THE  MAGNETIC  CIRCUIT  [ART.  67 

Uni-coil  windings  in  open  slots  ....      3  to    6  perms  per  cm. 
Thoroughly  distributed  windings  in 

open  slots 1.5  to    3      "          " 

Uni-coil  windings  in   completely 

closed  slots 7  to  14      "  " 

Thoroughly  distributed  windings  in 

completely  closed  slots 3  to    6      "  " 

The  much  larger  values  of  #>/  for  closed  slots,  as  compared 
to.  those  with  open  slots,  were  to  be  expected  because  the  bridge 
which  closes  the  slot  offers  a  path  of  high  permeance.  The 
lower  values  for  windings  distributed  in  several  slots  per  pole 
per  phase,  as  compared  to  uni-slot  windings,  are  due  to  the 
fact  that  the  partial  linkages  become  more  and  more  pronounced 
as  the  winding  is  distributed  into  a  larger  number  of  separate 
coils,  and  also  because  the  length  of  the  paths  is  greater.  This 
is  somewhat  analogous  to  splitting  a  transmission  line  into  two 
or  more  lines;  see  prob.  21  in  Art.  61.  The  greatest  reduction 
in  the  value  of  the  inductance  results  when  the  number  of  slots 
is  increased  from  one  to  two;  a  further  subdivision  is  of  much 
less  importance.  For  instance,  if  the  permeance  <P{  with  a 
uni-slot  coil  is  7,  then  dividing  the  same  coil  into  two  slots 
reduces  the  permeance  to  less  than  5.  On  the  other  hand,  a 
change  from  four  to  five  slots  per  phase  per  pole  would  hardly 
reduce  the  equivalent  permeance  more  than  from  say  3.5  to  3.4. 
The  data  in  the  table  above  give  rather  a  wide  range  from  which 
to  select  a  value  of  (Pj  for  a  particular  machine,  and  the  designer 
must  exercise  his  judgment  as  to  whether  his  machine  will  have 
a  permeance  nearer  the  upper  or  lower  limit.  This  judgment 
comes  with  experience,  by  comparing  the  predicted  performance 
of  machines  with  that  actually  observed. 

The  values  of  (Pa'  and  (Pe'  depend  upon  the  number  of  coils 
per  group,  in  other  words,  upon  the  number  of  slots  per  pole 
per  phase.  Until  more  accurate  and  detailed  data  are  available^ 

he  considers  separately  the  equivalent  permeance  (P8  of  each  slot,  instead 
of  the  group  of  slots  per  pole  per  phase.  Thus,  his  formula  for  the  leakage 
inductance  per  pole,  with  our  notation,  is  LPP  =  SPPCS2(PS,  where  Spp  is 
the  number  of  slots  per  pole  per  phase,  and  Cs  is  the  number  of  armature 
conductors  per  slot.  The  values  of  unit  permeance,  which  he  gives  and  denotes 
by  A,  refer  to  this  formula. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  231 

we  shall  assume  (Pe'  =  (Pa',  and  use  the  following  values,  based 
upon  Mr.  Hobart's  experiments:  J 

Number  of  slots  per  pole  per  phase  .1         2         3       more  than  3 
(Pef  =  <pa',  in  perms  per  centimeter  ..0.8     0.7     0.6  0.5 

With  a  fractional-pitch  winding  the  inductance  is  some- 
what reduced  (see  the  end  of  the  preceding  article).  As  an 
empirical  correction,  the  inductance  calculated  for  a  full-pitch 
winding  may  be  multiplied  by  the  winding  pitch  factor  kw. 

The  leakage  reactance  of  the  armature  cannot  be  calculated 
from  a  short-circuit  test,  because  the  short-circiut  current  is 
essentially  determined  by  the  direct  armature  reaction.  A 
difference  of  50  or  even  100  per  cent  in  the  armature  reactance 
would  change  the  short-circuit  current  by  only  a  few  per  cent. 
A  much  closer  approximation  is  obtained  from  the  so-called 
air-characteristic.2  Namely,  it  has  been  found  by  numerous  experi- 
ments that  the  armature  inductance,  when  the  field  is  revol- 
ving synchronously,  is  nearly  equal  to  the  armature  inductance 
with  the  field  completely  removed,  and  the  armature  supplied 
with  alternating  currents  from  an  external  source.  The  air- 
characteristic  is  the  relation  between  the  current  and  the  voltage 
under  these  conditions.  Eliminating  the  ohmic  drop,  the  induct- 
ance of  the  machine  is  easily  calculated,  and  the  value  so  found 
can  be  used  in  the  prediction  of  the  performance  of  the  machine, 
Such  an  air-characteristic  is  easily  taken  in  the  shop  or  in  the 
power  house  before  the  machine  is  completely  assembled.  From 
the  three  observed  curves,  namely,  the  no-load  saturation  curve, 
the  short-circuit  curve,  and  the  air-characteristic,  the  perform- 
ance of  a  synchronous  machine  at  any  load  can  be  predicted 
to  a  considerable  degree  of  accuracy. 

Prob.  18.  What  is  the  inductance  per  phase  of  a  6-pole,  3-phase, 
turbo-alternator,  the  armature  of  which  has  the  following  dimensions: 
Bore,  1.2  m.,  gross  length  of  core,  1.2  m.,  20  air-ducts  of  1  cm.  each, 
90  open  slots?  There  are  8  conductors  per  slot,  the  winding  is  of  the  two- 
layer  type,  the  winding  pitch  is  11/15.  Assume  (P/  =  2. 5  perms /cm. 

Ans.  24.3  mh. 

Prob.  19.  The  inductance  of  the  machine  specified  in  the  preceding 
problem  was  determined  experimentally  (from  -  an  air-characteristic), 

1  Journ.  Inst.  Electr.  Eng.  (British),  Vol.  31,  pp.  192  ff. 
2  Pichelmayer,  Dynamobau  (1908),  p.  207. 


232  THE  MAGNETIC  CIRCUIT  [ART.  68 

and  compared  to  that  measured  on  a  similar  machine,  the  gross  length 
of  the  core  of  which  was  80  cm.  and  which  was  provided  with  12  ducts 
of  1  cm.  each.  The  equivalent  leakage  permeance  of  the  shorter  machine 
was  found  to  be  30  per  cent  less  than  that  of  the  other  machine.  What 
are  the  actual  values  of  (P/  and  (Pe'(=(P</)  for  both  machines? 

Ans.     (P/  =  2.46,  OY=(P«'  =  0.56  perm/cm. 

Prob.  20.  An  alternator  has  3  slots  per  phase  per  pole,  and  the 
equivalent  permeance  is  (P/  =  1.8;  (Pe'  =  0.6.  What  would  be  the  value 
of  the  same  constants  per  slot?  Ans.  5.4  and  1.8. 

Prob.  21.  A  3-phase  alternator  has  4  slots  per  phase  per  pole. 
If  the  coils  were  connected  up  for  a  2-phase  machine  without  change 
what  would  be  the  ratio  of  the  new  L  to  the  old?  Ans.  3:2. 

68.  The  Reactance  Voltage  of  Coils  undergoing  Commuta- 
tion. Let  Fig.  57  represent  a  part  of  the  armature  winding 
and  commutator  of  a  direct-current  machine,  with  two  adjacent 
sets  of  brushes.  During  the  interval  of  time  when  an  arma- 
ture coil,  such  as  CD,  is  short-circuited  by  a  set  of  brushes, 
the  current  in  the  coil  is  reversed  from  its  full  value  in  one 
direction  to  an  equal  value  in  the  opposite  direction.  The  coil 
is  then  said  to  undergo  commutation. 

Under  unfavorable  conditions  this  reversal  of  current  is 
accompanied  by  sparking  between  one  of  the  edges  of  the  brushes 
and  the  commutator.  Unless  a  machine  is  provided  with  inter- 
poles,  its  output  is  usually  limited  by  this  sparking  at  the  com- 
mutator. It  is  of  importance,  therefore,  to  have  a  practical 
criterion  for  judging  the  quality  of  commutation  to  be  expected 
of  a  given  machine.  Numerous  formulae  and  methods  have 
been  proposed  for  the  purpose;  all  rational  formulae  contain, 
as  a  factor,  the  inductance  of  the  coils  undergoing  commutation, 
because  this  inductance  determines  essentially  the  law  according 
to  which  the  current  is  reversed  with  the  time.  For  this  reason, 
the  subject  of  commutation  is  treated  in  this  chapter,  under  the 
general  topic  of  the  inductance  of  windings.  The  method  of 
calculation  of  the  inductance  and  the  criterion  of  commutation 
given  below  are  due  to  Mr.  H.  M.  Hobart.1 

A  description  of  the  phenomenon  of  commutation.  The  phenom- 
enon of  commutation  may  be  briefly  described  as  follows: 
Let,  for  the  sake  of  explanation,  the  armature  and  the  commu- 

1  See  Hobart,  Elementary  Principles  of  Continuous-Current  Dynamo  Design 
(1906),  Chap.  4;  also  Parshall  and  Hobart,  Electric  Machine  Design  (1906), 
pp.  171-194. 


CHAP.  XII] 


INDUCTANCE  OF  WINDINGS 


233 


tator  be  assumed  to  be  stationary,  and  the  brushes  revolving  in 
the  direction  of  the  horizontal  arrow,  shown  in  Fig.  57.  Let 
the  machine  be  provided  with  a  multiple  winding  (lap  winding), 
so  that  there  are  as  many  armature  circuits  and  sets  of  brushes 
as  there  are  poles.  The  current  through  each  armature  branch 


is,  therefore 


(155) 


where  /  is  the  total  armature  current  and  p  the  number  of  poles. 


Commutator 


mim 

I      i     '  I  I        0 


nun 

Ml 

Toe^* 

1 

^HeelJ 

J 

Negative 
brush 


Direction  of  motion 
of  the  brushes 


FIG.  57. — Part  of  the    armature  winding,    commutator,  and  brushes  in  a 
direct  current  machine. 


At  each  set  of  brushes  two  branches  of  the  armature  winding 
are  connected  in  parallel,  so  that  the  current  through  each  set 
of  brushes  is  equal  to  2/x  With  reference  to  Fig.  57,  it  will 
be  seen  that  the  two  armature  branches,  X  and  Y,  which  begin 
at  each  set  of  brushes,  may  be  called,  with  respect  to  this  set, 
the  left-hand  branch  and  the  right-hand  branch. 


234  THE  MAGNETIC  CIRCUIT  [ART.  68 

In  the  position  of  the  positive  brushes  just  preceding  that 
marked  1,  the  coil  CD  is  not  short-circuited,  and  carries  the  full 
current  /i,  being  the  first  coil  of  the  right-hand  branch.  The 
lead  d  is  idle,  and  the  total  current  2/!  is  delivered  to  the  brushes 
through  the  leads  c,  m,  and  n.  In  the  position  of  the  positive 
brushes  just  after  that  marked  2,  the  coil  CD  is  again  not  short- 
circuited,  but  is  carrying  the  full  current  /i  in  the  opposite  direc- 
tion, being  the  first  coil  of  the  left-hand  branch. 

In  the  positions  of  the  brushes  between  1  and  2  the  coil  CD 
is  short-circuited  by  the  brushes  through  the  leads  c  and  d, 
and  the  current  in  the  coil  changes  gradually  from  +  /!  to  — /i. 
If  the  coil  possessed  no  inductance,  the  variation  in  the  current 
would  be  practically  determined  by  the  contact  resistance 
between  the  brush  and  the  commutator,  the  resistance  of  the 
coil  itself  and  of  the  leads  being  negligible  (with  carbon  brushes). 
Under  these  conditions  the  current  in  the  short-circuited  coil 
would  vary  with  the  time  according  to  the  straight-line  law, 
and  the  current  density  under  the  heels  and  the  toes  of  the 
brushes  would  be  the  same.  This  is  called  the  "  pure  resistance  " 
commutation,  or  the  perfect  commutaton,  because  it  is  not  accom- 
panied by  sparking.  Such  a  commutation  is  approached  in 
machines  with  inteupoles,  when  the  effect  of  the  inductance  is 
correctly  compensated  for  by  the  commutating  flux  (Art.  54). 

In  reality,  the  short-circuited  coil  possesses  a  considerable 
inductance,  which  has  the  effect  of  electromagnetic  inertia, 
retarding  the  reversal  of  the  current.  Consequently,  at  the 
beginning  of  the  commutation  period  the  lead  d  and  the  corre- 
sponding commutator  segment  do  not  carry  their  proper  share 
of  the  current,  which  they  would  carry  with  a  perfect  com- 
mutation. At  the  end  of  the  commutation  period  the  current 
must  then  be  reversed  quickly,  because  the  whole  current  must 
be  transferred  from  the  lead  c  to  the  other  leads.  If  the  inductance 
is  considerable,  the  current  in  the  lead  c  is  still  of  a  considerable 
magnitude  when  the  toe  of  the  brush  is  about  to  leave  the 
corresponding  commutator  segment.  Therefore,  the  last  period 
of  the  reversal  is  accomplished  through  the  air  between  the 
brush  and  the  segment,  in  the  form  of  an  electric  arc.  This 
is  known  as  the  sparking  at  the  brushes.  Besides,  during  the 
last  moments  of  reversal,  the  current  density  under  the  toe 
is  much  higher  than  the  average  density  under  the  brush,  and 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  235 

this  high  density  causes  a  glowing  at  the  edge  of  the  brushes, 
making  the  commutation  still  less  satisfactory. 

The  average  reactance  e.m.f.  in  the  coils  of  a  full-pitch  lap 
winding.  For  an  empirical  criterion  of  the  quality  of  commutation 
Mr.  Hobart  takes  the  average  reactance  voltage  induced  in  the  coil. 
This  is  a  reasonable  criterion,  because  the  ratio  of  the  maximum 
voltage  occurring  when  the  brush  leaves  a  segment  to  the  average 
voltage,  will  be  more  or  less  the  same  in  machines  of  usual 
design  constants.  Of  course,  the  average  reactance  voltage 
is  only  a  relative  criterion,  to  be  used  with  great  discretion, 
and  applied  only  for  comparison  with  machines  which  proved 
in  actual  operation  to  commutate  satisfactorily. 

Let  the  inductance  of  an  armature  coil  between  two  adjacent 
commutator  segments  be  Leq.  The  subscript  eq  (meaning  equiva- 
lent) is  added  to  indicate  that  the  value  of  L  includes  not  only  the 
true  inductance  of  the  coil  itself,  but  also  the  average  inductive 
action  of  the  coils  which  are  undergoing  commutation  simultane- 
ously with  it.  Let  the  frequency  of  the  current  in  the  coil  under- 
going commutation  be  /  cycles  per  second.  Then  the  current  is 
reversed  in  a  time  J/.  The  flux  changes  during  this  time 
from  +LeQIi  to  —Leqli:  Hence,  according  to  the  fundamental 
eq.  (26)  Art.  24,  the  average  reactance  voltage,  which  is  taken 
as  the  criterion  of  commutation,  is 


In  order  to  obtain  a  satisfactory  commutation,  the  voltage 
eave  must  not  exceed  a  certain  value,  determined  from  actual 
experience  with  machines  in  regular  operation.  Mr.  Hobart 
recommends  values  for  eave  not  to  exceed  3  to  4  volts,  provided 
that  one  does  not  depend  upon  the  fringe  flux  of  the  main  poles 
to  facilitate  commutation. 

The  inductance  Leq,  which  enters  in  the  foregoing  formula, 
is  calculated  according  to  the  general  formula  (150),  as  follows: 
Assume  first  that  there  is  no  common  flux  or  mutual  induction 
between  the  coil  under  consideration  and  the  other  coils  which 
are  simultaneously  short-circuited.  Then,  if  q  is  the  number 
of  turns  per  commutator  segment  (in  Fig.  57  #=1)  we  must  put 
CPP=q.  This  will  give  the  inductance  of  one  side  of  the  coil, 
say  C.  To  obtain  the  inductance  of  both  sides,  C  and  D,  the 
result  must  be  multiplied  by  2,  or  Leq=2LPP. 


236  THE  MAGNETIC  CIRCUIT  [ART.  68 

In  reality  there  is  a  common  flux  which  links  with  the  coil 
under  consideration  and  with  the  other  coils  undergoing  com- 
mutation at  the  same  time.  This  flux  is  changing  with  the 
time,  and  consequently  it  induces  additional  voltages  in  the 
coil  CD.  The  induced  e.m.f.  depends  upon  the  relative 
position  of  CD  and  the  other  coils  (whether  in  the  same  slot 
or  in  the  adjacent  slots)  and  upon  the  rate  of  change  of  the  cur- 
rent in  each  coil  of  the  group.  It  would  be  too  complicated  for 
practical  purposes  to  take  all  these  factors  into  account  with 
any  degree  of  accuracy.  Therefore,  Hobart  makes  a  further 
assumption,  namely,  that  the  current  in  all  the  coils,  which  are 
short-circuited  at  the  same  time,  varies  at  the  same  rate  and  that 
the  whole  leakage  flux  is  linked  with  all  the  coils  of  the  group  (Fig. 
57). 

Let  s  be  the  average  number  of  coils  simultaneously  short- 
circuited  under  a  set  of  brushes  (the  actual  number  varies  from 
instant  to  instant).  Consider  a  group  of  mutually  influencing 
conductors,  such  as  are  shown  at  C  or  at  D.  One-half  of  the 
conductors  in  the  same  group  are  short-circuited  by  the  positive 
brushes,  the  other  half  by  the  adjacent  negative  brushes.  The 
total  number  of  coils  in  each  group  is  2s,  and  since  by  assumption 
the  current  in  all  of  them  varies  at  the  same  rate,  and  all  of 
the  flux  is  linked  with  all  of  the  coils,  the  equivalent  inductance 
of  the  coil  A B  is  2s  times  larger  than  if  this  coil  were  alone. 
Thus  for  a  multiple-wound  armature 

L^ = 2LPP .  2s = 4sq2  ((Pfli  +  (Pa'la + %(P >e'le)  X  10~8  henrys.     (157) 

On  the  basis  of  Mr.  Hobart 's  tests  and  until  more  accurate  data 
are  available,  the  following  average  values  of  the  unit  permeances 
may  be  used:  (Pj  =  4  and  (Pa'  =  <Pe'  =  0.8  perms  per  centimeter. 

The  frequency  /  which  enters  into  formula  (156)  is  calculated 
as  follows :  The  time  between  the  positions  1  and  2  of  the  brushes 
corresponds  to  one-half  of  one  cycle,  because  during  this  interval 
the  current  changes  from  +/i  to  —I\.  Let  v  be  the  peripheral 
velocity  of  the  commutator,  in  meters  per  sec.,  let  b  be  the  thick- 
ness of  the  brushes,  and  b'  the  thickness  of  the  mica  insulation 
between  the  commutator  segments,  both  in  millimeters.  The 
time  between  the  positions  1  and  2  of  the  brushes  is  (6  —  bf)/WOOv 
seconds.  Hence 

f=5QOv/(b-V)  cy./sec (158) 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  237 

The  number  of  simultaneously  short-circuited  coils  varies 
periodically  with  the  position  of  the  brushes.  Thus,  in  Fig.  57 
sometimes  two  and  sometimes  three  coils  are  short-circuited 
by  one  set  of  brushes.  On  the  average 

s=(b-b')/a,       .    .    .    .    .    .     (159) 

where  a  is  the  width  of  one  commutator  segment  including  the 
mica  insulation.  Thus,  all  the  values  which  enter  into  the 
formula  (156)  are  determined,  and  the  reactance  voltage  for  a 
given  machine  can  be  easily  calculated. 

Formula  (156)  is  used  not  only  as  a  criterion  of  the  commuta- 
tion, but  also  for  the  calculation  of  the  flux  density,  under  the 
interpoles  where  such  are  required.  Namely,  this  flux  density  must 
be  such  that  the  average  voltage  induced  by  the  commutating 
flux  is  approximately  equal  and  opposite  to  the  average  reactance 
voltage;  see  Art.  24.  prob.  6.  A  still  closer  compensation  for  the 
influence  of  the  inductance  is  achieved  by  properly  grading  the 
commutating  flux,  so  as  to  compensate  not  only  for  the  average 
reactance  voltage,  but  also  to  some  extent  for  the  instantaneous 
induced  e.m.fs. 

The  average  reactance  e.m.f.  induced  in  some  other  direct  cur- 
rent windings.  With  two-circuit  wave  windings  two  cases  must 
be  considered,  namely,  (a)  when  the  machine  is  provided  with 
only  two  sets  of  brushes,  (b)  when  there  are  more  than  two  sets 
of  brushes.  In  the  first  case  eq.  (156)  is  used,  where 


and  Leq  is  understood  to  comprise  the  short-circuited  conductors 
under  all  the  poles,  per  commutator  segment.  Let  there  be 
again  q  turns  per  coil,  that  is,  per  unit  of  winding  per  pair  of 
poles.  Since  the  corresponding  conductors  under  all  the  poles 
are  in  series,  we  have  that  Leq=pLPP.  The  influence  of  the 
other  simultaneously  short-circuitqd  coils  is  expressed  as  before 
by  the  factor  2s,  where  s  is  given  by  formula  (159).  Thus,  for 
a  two-circuit  winding  with  two  sets  of  brushes 

Leg=2psq2((Pi'li+(Pa'la  +  i(Pe'le)XlO-1  henrys.      .     (161) 

The  frequency  /is  given  as  before  by  eq.  (158). 

When  more  than  two  sets  of  brushes  are  used,  the  sets  of 
equal  polarity  are  connected  in  parallel  by  the  stud  connections 


238  THE  MAGNETIC  CIRCUIT  [ART.  68 

outside  the  armature  windings,  and  beside  there  are  two  short- 
circuiting  paths  through  the  coils  undergoing  commutation:  a 
long  path  and  a  short  path.  Thus,  the  problem  becomes 
indefinite,  because  it  is  not  possible  to  tell  the  relative  amounts  of 
the  current  through  these  different  paths.  Disregarding  the  short 
path,  the  criterion  becomes  the  same  as  in  the  case  of  a  machine 
with  two  sets  of  brushes  only.1  This  is  on  the  safe  side,  and 
the  commutation  may  be  expected  to  be  better  than  that  cal- 
culated, or  at  the  worst,  as  good. 

With  multiplex  windings,  the  expression  for  eave  is  the 
same  as  that  given  above,  provided  that  proper  values  are  selected 
for  /!,  s,  and  /.  With  regard  to  the- latter  quantity  it  must 
be  remembered  that  bf  is  much  larger  than  the  actual  thickness 
of  mica.  •  Namely,  with  respect  to  the  component  winding  under 
consideration  the  metal  of  the  commutator  segments  belonging 
to  the  other  component  windings  is  equivalent  to  insulation. 
This  fact  must  not  be  lost  sight  of  in  choosing  the  correct  value 
for  bf  to  be  used  in  the  expression  (158). 

With  fractional-pitch  windings  the  reactance  voltage  is  smaller 
than  with  the  corresponding  full-pitch  winding,  because  the 
conductors  short-circuited  under  the  adjacent  sets  of  brushes 
(Fig.  57)  are  situated  in  part  or  totally  in  different  slots,  and 
have  a  smaller  common  magnetic  flux,  or  none  at  all.  When 
the  winding  pitch  is  reduced  considerably,  s  instead  of  2s  must 
be  used  in  the  preceding  formulae;  otherwise  a  value  between 
s  and  2s  must  be  chosen,  according  to  one's  judgment.2 

Prob.  22.  The  armature  of  a  6-pole,  600-r.p.m.,  multiple-wound, 
direct-current  machine  has  the  following  dimensions:  Diameter,  85  cm.; 
gross  length,  22  cm.;  three  air-ducts,  1  cm.  each;  1008  face  conductors. 

1  For  an  analysis  of  this  case  see  C.  A.  Adams,  Reactance  E.M.F.  and 
the  Design  of  Commutating  Machines,  Electrical  World  and  Engineer,    Vol. 
46  (1905),  p.  346. 

2  For  an  advanced  and  more  scientific  theory  of  commutation,  see  Arnold, 
Die  Gleichstromaschine,  Vol.  1   (1906),  pp.  354   to   513;    in  particular  the 
approximate  formula  (170)  on  bottom  of  p.  498;  also  Vol.  2  (1907),  chapter 
14.     A  simpler  and  more  concise  treatment  will  be   found   in   Tomalen's 
Electrical  Engineering.     A  good  practical  treatment  will  also  be  found  in 
Pichelmayer's   Dynamobau,   pp.    86-118;    it   is   considerably   simplified   as 
compared  to  Arnold's  treatment,  and  is  accurate  enough  for  practical  pur- 
poses, because  the  numerical  values  of  unit  permeances  are  known  only 
approximately. 


CHAP.  XII]  INDUCTANCE  OF  WINDINGS  239 

The  commutator  diameter  is  52  cm.;  the  number  of  segments,  252; 
the  mica  insulation  is  1  mm.  thick;  the  brushes  are  15  mm.  thick. 
What  is  the  average  reactance  voltage  when  the  total  armature  current 
is  320  amp.?  Ans.  4.52  volts. 

Prob.  23.  Show  that  the  answer  to  the  preceding  problem  would 
be  nine  times  larger  if  by  mistake  the  winding  were  assumed  to  be  of 
the  two-circuit  type. 

Prob.  24.  The  peripheral  velocity  of  a  commutator  is  18  meters 
per  sec.,  the  width  of  each  segment  (without  mica)  is  4.5  mm.;  the 
thickness  of  the  mica  is  0.9  mm.  The  commutator  is  to  be  used  in  con- 
nection with  a  duplex  winding.  What  is  the  smallest  permissible 
thickness  of  the  brushes  if  the  frequency  of  commutation  must  not 
exceed  800  cycles  per  sec.?  Ans.  17.5  mm. 

Prob.  25.  For  a  perfect  commutation  and  for  an  imperfect  one 
draw  the  following  curves  to  time  as  abscissae:  (a)  the  current  in  the 
short-circuited  coil;  (6)  the  currents  in  the  leads  c  and  d]  (c)  the 
current  densities  under  the  heel  and  toe  of  the  brush.  Take  the  width 
of  the  brushes  to  be  equal  to  that  of  one  commutator  segment,  and 
assume  the  mica  insulation  to  be  of  a  negligible  thickness. 

Prob.  26.  Show  that  the  width  of  the  brushes  has  comparatively 
little  net  effect  upon  the  commutation  of  a  machine. 

Prob.  27.  What  flux  density  is  needed  in  the  interpolar  zone  in 
prob.  22  to  secure  perfect  commutation?  Ans.  2.23  kl./sq.  cm. 


CHAPTER  XIII 

THE  MECHANICAL  FORCE  AND  TORQUE  DUE  TO 
ELECTROMAGNETIC    ENERGY. 

69.  The  Density  of  Energy  in  a  Magnetic  Field.  The  reader 
is  already  familiar  with  the  fact  that  a  certain  amount  of  energy 
is  required  to  establish  the  flux  within  a  magnetic  circuit,  and 
that  this  energy  remains  stored  in  the  field.  This  stored  energy 
may  be  conveniently  thought  of  as  the  kinetic  energy  of  vor- 
tices around  the  lines  of  force  (Art.  3).  Various  expressions 
for  the  total  stored  electro-magnetic  energy  are  given  in  Arts. 
57  and  58;  the  problem  here  is  to  find  a  relation  between  the 
distribution  of  the  flux  density  and  that  of  the  energy  in  the 
field. 

Consider  first  the  simplest  magnetic  circuit  (Fig.  1)  con- 
sisting of  a  non-magnetic  material.  According  to  the  last  eq. 
(99),  the  total  energy  stored  in  such  a  circuit  is 


joules,    .     .     .    .  :.     (162) 

if  0  is  in  webers,  I  and  A  in  cm.,  and  /*=  1.257X10"8  henrys 
per  cm.  cube.  The  volume  of  the  field  is  V=IA  cubic  cm. 
Since  the  flux  density  is  uniform,  the  energy  is  also  uniformly 
distributed,  and  the  density  of  the  energy  is 


Denoting  the  density  of  the  energy  W/V  by  W,  and  introducing 
the  flux  density  B=  4>/A,  we  get 

TF'  =  fB2//*  joules  per  cu.cm.   .     .     .     (163) 

Either  B  or  //  can  be  eliminated  from  this  expression  by  means 
of  the  relation  B  =  pH,  so  that  we  have  two  other  expressions 
for  the  density  of  the  energy: 

.......     (164) 

(165) 
240 


CHAP.  XIII]        TORQUE  AND  TRACTIVE  EFFORT  241 

Two  more  expressions  for  the  density  of  the  energy  can  be  written, 
using  the  reluctivity  v  instead  of  the  permeability  /*. 

In  a  uniform  field  the  preceding  expressions  represent  the 
actual  amounts  of  energy  stored  per  cubic  centimeter.  In  a 
non-uniform  field  W  is  the  density  of  energy  at  a  point,  or  the 
limit  of  the  expression  AW/AV.  This  is  analogous  to  what  we 
have  in  the  case  of  a  non-uniform  distribution  of  matter,  where 
the  density  of  matter  at  a  point  is  the  limit  of  the  ratio  of  the 
mass  to  the  volume.  Thus,  the  total  energy  stored  in  a  non- 
uniform  field  is 


(166) 


where  the  integration  is  to  be  extended  over  the  volume  of  the 
whole  magnetic  circuit.  Similarly,  from  eqs.  (164)  and  (165) 
we  get 


flr-i/il     H2dV, (167) 

W=%C  HBdV (168) 


These  expressions  are  consistent  with  eqs.  (102)  and  (102a) 
as  is  shown  in  pfob.  6  below. 

When  fj.  is  variable,  the  preceding  formulae  do  not  hold  true, 
and  the  density  of  energy  is  represented  by  eq.  (19),  Art.  16. 

Prob.  1.  Deduce  an  expression  for  the  magnetic  energy  stored  in 
the  insulation  of  a  concentric  cable  (Fig.  46),  between  the  radii  a  and  b, 
the  length  of  the  cable  being  I  cm.  and  the  current  i.  Hint:  For  an 
infinitesimal  shell  of  a  radius  x  and  thicknesss  dx  we  have :  H  =  i/2nx, 
and  dV=2nxl  dx.  Ans.  W  =  0.23ft2  log  (6/o)10-8  joules. 

Prob.  2.  Check  the  answer  to  the  preceding  problem  by  means 
of  eqs.  (104)  and  (109). 

Prob.  3.  In  a  concentric  cable  (Fig.  46)  a  =  7  mm.  and  b  is  20  mm. 
What  is  the  density  of  the  energy  at  the  inner  and  outer  conductors, 
when  i  =  120  amp.? 

Ans.    4.68  and  0.57  microjoules  per  cu.cm. 

Prob.  4.     Deduce  expression  (110)  from  eq.  (167). 

Prob.  5.  Taking  the  data  from  the  various  problems  given  in  this 
book  as  typical,  show  that  ordinarily  in  generators  and  motors  a  large 
proportion  of  the  total  energy  of  the  field  is  stored  in  the  air-gap. 

Prob.  6.  Show  that  eqs.  (166)  to  (168)  are  consistent  with  eqs. 
(102)  to  (103u).  Solution :  Take  an  infinitesimal  tube  of  partial  linkages 


242  THE  MAGNETIC  CIRCUIT  [ART.  70 

(Fig.  45).  The  energy  contained  in  this  tube  is  dW  =  %MPd(I>p;  but 
MP=  I  Hdl,  and  d$=BdA.  Since  d@  is  the  same  through  all  cross- 
sections  of  the  tube,  d@  can  be  introduced  under  the  integral  sign,  and 
we  have  dW '  =  i  \ HdlBdA  =J  f HBdV,  the  integration  being  extended 

over  the  volume  of  the  tube.  The  total  energy  of  the  circuit  is  found 
by  extending  the  integration  over  the  volume  of  all  the  tubes  of  the 
field.  The  other  equations  are  proved  in  a  similar  manner. 

70.  The  Longitudinal  Tension  and  the  Lateral  Compression 
in  a  Magnetic  Field.  The  existence  of  mechanical  forces  in  a 
magnetic  field  is  well  known  to  the  student.  He  needs  only 
to  be  reminded  of  the  supporting  force  of  an  electromagnet,  of 
the  attraction  and  repulsion  between  parallel  conductors  carrying 
electric  currents,  of  the  torque  of  an  electric  motor,  etc.  These 
mechanical  forces  must  necessarily  exist,  if  the  magnetic  field 
is  the  seat  of  stored  energy.  This  is  because,  if  we  deform  the 
circuit,  we  must  in  general  change  the  stored  energy  and  hence 
do  mechanical  work.  The  lines  of  force  tend  to  shorten  them- 
selves and  to  spread  laterally,  so  as  to  make  the  permeance  of 
the  field  a  maximum,  with  the  complete  linkages.  Where  there 
are  partial  linkages,  it  is  the  total  stored  energy  that  tends  toward 
a  maximum  (Art.  57).  This  fact  is  entirely  consistent  with 
the  hypothesis  of  whirling  tubes  of  force,  because  the  centrifugal 
force  of  rotation  produces  exactly  the  same  effect,  that  is,  a  lateral 
spreading  and  a  tension  along  the  axis  of  rotation.  A  good 
analogy  is  afforded  by  a  short  piece  of  rubber  tube  filled  with 
water  and  rotated  about  its  longitudinal  axis. 

(a)  The  Longitudinal  Tension.  Consider  again  the  simple 
magnetic  circuit  (Fig.  1),  and  let  it  be  allowed  to  shrink,  due 
to  the  longitudinal  tension  of  the  lines  of  force,  so  as  to  reduce 
its  average  length  by  Al,  without  changing  the  cross-section  A. 
Let  at  the  same  time  the  current  be  slightly  decreased  so  as  to 
keep  the  same  total  flux  as  before.  Let  Ftf  be  the  mechanical 
tension  along  the  lines  of  force,  per  square  cm.  of  cross-section 
A;  then  the  mechanical  work  done  against  the  external  forces 
which  hold  the  winding  stretched  is  (Ft'.A)  Al.  The  density 
of  energy  W  remains  the  same  because  B  is  the  same,  but  the 
total  stored  energy  is  decreased  by  W'(AM),  because  the  volume 
of  the  field  is  decreased  by  AM.  Since  the  change  was  made 


CHAP.  XIII]         TORQUE  AND  TRACTIVE  EFFORT 


243 


in  such  a  way  as  to  keep  the  total  flux  constant,  no  e.m.f.  was 
induced  in  the  winding  during  the  deformation,  and  consequently 
there  was  no  interchange  of  energy  between  the  electric  and 
the  magnetic  circuit.  Thus,  the  decrease  in  the  stored  energy 


Coil 


FIG.  58.  —  A  lifting  electromagnet. 

is  due  entirely  to  the  mechanical  work  performed. 
the  two  preceding  expressions,  we  have  that 


Equating 


.    .,{.    .     (169) 

If  W  is  in  joules  per  cu.cm.,  F't  is  in  joulecens  per  sq.cm.  (see 
Appendix  I)  ,  so  that  in  a  rational  system  of  units  the  mechanical 
stress  per  unit  area  is  numerically  equal  to  the  density  of  the  stored 


244      |;|  THE  MAGNETIC  CIRCUIT  [ART.  70 

energy.    The  physical  dimensions  of  F'  and   W  are  also  the 
same. 

If  Ft  is  in  kg.  per  sq.cm.,  B  in  kilolines  per  sq.cm.,  and 
H  in  kiloampere-turns  per  cm.,  the  preceding  formula  becomes, 
when  applied  to  air, 


(170) 


These  formulae  apply  directly  to  the  lifting  magnet  (Fig. 
58),  and  give  the  carrying  weight  per  unit  area  of  the  contact 
between  the  core  and  its  armature.  The  total  weight  which 
the  magnet  is  able  to  support  is 


where  A  is  the  sum  of  the  areas  denoted  by  Si  and  82.  Of  course, 
H  is  taken  for  the  air-gap,  which  is  the  only  part  of  the  circuit 
that  is  changing  its  dimensions  when  the  armature  is  moved. 

(b)  The  Lateral  Compression.  Let  now  the  simple  magnetic 
circuit  be  allowed  to  expand  laterally  by  a  small  length  Js  in 
directions  perpendicular  to  the  surface  of  the  toroid.  Let  Fc' 
be  the  pressure  (compression)  exerted  by  the  lines  of  force  upon 
the  winding,  per  sq.  cm.  of  the  surface  of  the  toroid.  Then 
the  mechanical  work  done  by  the  magnetic  forces  in  expanding 
the  ring  against  the  external  forces  which  hold  the  winding, 
is  SFC'AS,  where  S  is  the  surface  of  the  toroid.  Let  again  the 
current  be  slightly  decreased  during  the  deformation,  so  as  to 
keep  the  flux  constant.  No  voltage  is  induced  in  the  winding, 
and  hence  there  is  no  interchange  of  energy  between  the  electric 
and  the  magnetic  circuit.  Thus  we  can  find  Fc',  as  we  found 
the  stress  in  the  case  of  the  tension,  by  equating  the  work  done 
to  the  decrease  in  the  stored  energy.  The  stored  energy  is 
expressed  by  eq.  (162),  in  which  A  is  the  only  variable;  hence 
by  differentiating  W  with  respect  to  A  we  get  : 


This  is  a  negative  quantity,  because  the  stored  energy  decreases. 
But  IAA  represents  the  increase  in  the  volume  of  the  ring,  so 
that  IAA  =  SJs,  and  consequently 

Fe'  =  l&/n=ti,H*  =  W'~Ft'  .....     (172) 
In  other  words,  the  lateral  compression  is  nummerically  equal  to 


CHAP.  XIII]         TORQUE  AND  TRACTIVE  EFFORT 


245 


the  longitudinal  tension,  and  both  are  numerically  equal  to  the 
density  of  the  stored  energy. 

As  an  application  of  the  lateral  action,  consider  a  constant- 
current  or  floating-coil  transformer  (Fig.  59),  used  in  series 
arc-lighting.  The  leakage  flux  is  similar  in  its  character  to  that 
shown  in  Figs.  50  and  51.  The  lateral  pressure  of  the  leakage 
lines  between  the  coils  tends  to  separate  them,  acting  against 
the  weight  of  the  floating  coil.  A  part  of  this  weight  has  to  be 
balanced  by  a  counter-weight  Q  because  the  electro-magnetic 
forces  under  normal  operation  are  comparatively  small. 

Since  the  currents  are  alternating,  the  force  is  pulsating, 


_T                    I/J                           /Core 

Hi 

j| 

> 

1 
1 

I 

1 

.            Tn  Constant* 
/                Current 
'/—  —  Series  A.C. 
s                  Lamps 

Counter- 
^  weight 

^ 

Secondary 
Goil 

i5 



H- 
v^ 

|t*             *       'I 

From  a  Constant 
Potential  A.C. 
Supply 

T 

Primary  Goil/ 

FIG.  59. — A  floating-coil  constant-current  transformer. 

but  is  always  in  the  same  direction,  tending  to  separate  the  coils. 
The  average  force  depends  upon  the  average  value  of  H2t  in 
other  words,  upon  the  effective  value  of  the  current.  According 
to  eqs.  (170)  and  (172),  we  have 

(Fc)ave=(Fc')ave.S=HeffS/15.Qkg.,       .    ..   (173) 

where  S  is  the  area  of  the  floating  coil  in  contact  with  the  flux. 
With  the  assumed  paths  for  the  lines  of  force,  and  neglecting 
the  reluctance  of  the  iron  core,  we  have  that 

Heff=nIeff/WQOl  kiloamp.-turns/cm.,     .    .     (174) 

where  I  is  the  length  of  the  lines  of  force  in  *  the  air,  in  cm.,  and 
nleff  is  the  m.m.f.  of  either  coil.  The  force  of  repulsion  is  pro- 
portional to  the  square  of  the  current,  and  is  independent  of  the 


246  THE  MAGNETIC  CIRCUIT  [ART.  70 

distance  h  between  the  coils.  Hence,  a  constant  weight  Q 
regulates  for  a  constant  current.  When  the  coils  are  further 
from  each  other,  the  induced  secondary  voltage  is  less,  on  account 
of  a  much  higher  leakage  flux.  When  the  current  increases 
momentarily,  due  to  a  decreasing  line  resistance,  the  coil  is 
overbalanced  and  rises  till  the  induced  voltage  and  current  fall 
to  the  proper  value.  Thus,  the  coil  always  floats  at  the  proper 
height  to  induce  the  voltage  needed  on  the  line. 

Formulae  (173)  and  (174)  apply  also  to  the  mechanical  forces 
between  the  primary  and  the  secondary  coils  of  a  constant- 
potential  transformer  (Figs.  13  and  51).  Under  normal  con- 
ditions these  forces  are  negligible,  but  in  a  violent  short-circuit 
the  end-coils  are  sometimes  bent  away  and  damaged,  unless 
they  are  properly  secured  to  the  rest  of  the  winding.  Such 
short-circuits  are  particularly  detrimental  in  large  transformers, 
having  a  close  regulation,  that  is,  having  a  very  small  internal 
impedance  drop,  and  which  are  connected  to  systems  of  practically 
unlimited  power  and  constant  potential.  As  Dr.  Steinmetz  puts 
it,  the  closest  approach  to  the  appearance  of  such  a  transformer 
after  a  short-circuit  is  the  way  two  express  trains  must  look 
after  a  head-on  collision  at  high  speed. 

Another  interesting  example  of  the  effect  of  the  mechanical 
forces  produced  by  a  magnetic  field  is  the  so-called  pinch  phe- 
nomenon.1 The  lines  of  force  which  surround  a  cylindrical  con- 
ductor may  be  compared  to  rubber  bands,  which  tend  to  com- 
press it.  With  a  liquid  conductor  and  large  currents,  such 
for  instance  as  are  carried  by  a  molten  metal  in  some  electro- 
metallurgical  processes,  the  pressure  of  the  magnetic  field  is 
sufficient  to  modify  and  to  reduce  the  cross-section  of  the  liquid 
conductor.  This  was  first  observed  by  Mr.  Carl  Hering  and 
called  by  him  the  pinch  phenomenon.  In  passing  a  relatively 
large  alternating  current  through  a  non-electrolytic  liquid  con- 
ductor contained  in  a  trough,  he  found  that  the  liquid  contracted 
in  cross-section  and  flowed  up-hill  lengthwise  in  the  trough, 
climbing  up  on  the  electrodes.  With  a  further  increase  of 

1  E.  F.  Northrup,  Some  Newly  Observed  Manifestations  of  Forces  in  the 
Interior  of  an  Electric  Conductor,  Physical  Review,  Vol.  24  (1907),  p. 
474.  This  article  contains  some  cleverly  devised  experiments  illustrating 
the  pinch  phenomenon,  and  also  a  mathematical  theory  of  the  forces  which 
come  into  play. 


CHAP.  XIII]         TORQUE  AND  TRACTIVE  EFFORT 


247 


current,  this  contraction  of  cross-section  became  so  great  at 
one  point  that  a  deep  depression  was  formed  in  the  liquid,  with 
steeply  inclined  sides,  like  the  letter  V. 

In  most  cases  of  mechanical  forces  in  a  magnetic  field,  these 
forces  and  the  resulting  movements  are  due  to  the  combined 


Stop 


Coil 


Core 


FIG.  60. — A  tractive  electromagnet.          FIG.  61.- 


-Two  bus-bars  and  their 
support. 


action  of  longitudinal  tensions  and  transverse  compressions  and 
not  to  one  of  these  actions  alone.  For-  instance,  a  loop  of  flexible 
wire,  through  which  a  large  current  is  flowing,  tends  to  stretch 
itself  so  as  to  assume  a  maximum  opening,  that  is,  a  maximum 
permeance  of  the  magnetic  field  linked  with  it.  This  action 
is  due  to  both  the  longitudinal  tension  and  the  lateral  pressure. 
In  such  cases  the  mechanical  forces  are  best  computed  by  the 
principle  of  virtual  displacements  explained  in  the  next  article. 


248 


THE  MAGNETIC  CIRCUIT 


[ART.  70 


Prob.  7.  Show  that  the  required  flux  density  in  the  air-gap  of  a 
lifting  electromagnet  (Fig.  58)  can  be  calculated  from  the  expression 
B  =  15.7VoF/A,  in  kl/sq.cm.,  where  F  is  the  rated  supporting  force, 
in  metric  tons,  A  is  the  area  of  contact  in  sq.  dm.,  and  a  is  the  factor 
of  safety. 

Prob.  8.  Show  that  in  an  armored  tractive  magnet  (Fig.  60)  the 
tractive  effort  F  varies  with  the  air-gap  s  according  to  the  law  Fsz  =  3.G8 
kg-cm.,  when  the  excitation  is  2000  amp.-turns  and  the  cross-section 
of  the  plunger  and  of  the  stop  is  12  sq.cm.  Assume  the  leakage  and  the 
reluctance  of  the  steel  parts  to  be  negligible. 

Prob.  9.  Referring  to  the  preceding  problem,  what  is  the  true 
average  pull  between  the  values  s  =  l  and  s=4  cm.,  and  what  is  the 
arithmetical  mean  pull?  Ans.  0.77  and  1.63  kg. 

Prob.  10.  Indicate  roughly  the  principal  paths  of  magnetic  leakage 
in  Fig.  60,  and  explain  the  influence  of  the  leakage  upon  the  tractive 
effort,  with  a  small  and  a  large  air-gap. 

Prob.  11.  The  flux  between  two  thin  and  high  bus-bars,  placed  at 
a  short  distance  from  each  other,  has  the  general  character  shown  in 
Fig.  61.  Calculate  the  force  per  meter  length  that  pushes  the  bus-bars 
apart  when,  during  a  short-circuit,  the  estimated  current  is  50  kilo- 
amperes.  Ans.  About  800  kg.  per  meter. 

Prob.  12.  Deduce  an  expression  for  the  magnetic  pull  due  to  the 
eccentric  position  of  the  armature  in  an  electric  machine  (Fig.  62). 
A  certain  allowance  is  usually  made  for  this  pull  in  addition  to  the 
weight  of  the  revolving  part,  in  determining  the  safe  size  of  the  shaft. 

Solution :  Since  the  pull  is  proportional 
to  the  square  of  the  flux  density,  we 
replace  the  actual  variable  air-gap 
density  by  a  constant  radial  density 
acting  upon  the  whole  periphery  of  the 
armature  and  equal  to  the  quadratic 
average  (the  effective  value)  of  the 
actual  flux  density  distribution.  Let 
this  value  be  Beff  kl.per  sq.  cm.  when 
the  armature  is  properly  centered. 
Let  the  original  uniform  air-gap  be  a, 
and  the  eccentricity  be  s.  Since  a  and 
e  are  small  as  compared  to  the  diameter 
of  the  armature,  the  actual  air-gap 
at  an  angle  a  from  the  vertical  is 
approximately  equal  to  a  —  ecosa.  Neg- 
lecting the  reluctance  of  the  iron  parts 

of  the  machine,  the  flux  density  is  inversely  as  the  length  of  the  air- 
gap,  so  that  we  have 

Ba=Beffa/(a-£  cos  a)  =Beff/[l  -  (e/o)  cos  a]. 


a— e 
FIG.  62. — An  eccentric  armature- 


Let  A  be  the  total  air-gap  area  to  which  B  refers;  then,  according 


CHAP.  XIII]        TORQUE  AND  TRACTIVE  EFFORT  249 

to  eq.  (171),  the  vertical  component  of  the  pull  upon  the  strip  of  the 
width  dais 


a>cos  a. 

The  horizontal  component  of  the  pull  is  balanced  by  the  corresponding 
component  on  the  other  half  of  the  armature.  The  total  pull  downward  is 

/•* 

Fex=[2A/(27tX24.7)]  I     Ba2  cos  a  da. 
Jo 

Putting  tan  a  =z  and  integrating  we  get  the  so-called  Sumec  formula 
for  the  eccentric  pull,  in  kg.  : 

Fex  =  (ABeff*/24.7)(e/a)[l-(s/ay]-™.    .    .    .     (175) 

The  integration  is  simplified;  if,  before  integrating,  the  difference  is 
taken  between  the  vertical  forces  at  the  points  corresponding  to 
a  and  to  n  —  a.  The  limits  of  integration  are  then  0  and  %x.1 

Prob.  13.  The  average  flux  density  under  the  poles  of  a  direct- 
current  machine  is  6.5  kl/sq.cm.;  the  poles  cover  68  per  cent  of  the 
periphery.  The  diameter  of  the  armature  is  1.52  m.;  the  effective 
length  is  56  cm.  What  is  the  magnetic  pull  when  the  eccentricity  is 
10  per  cent  and  when  it  is  50  per  cent  of  the  original  air-gap? 

Ans.  3.2  and  24  metric  tons. 

Prob.  14.  The  22/2-kv.,  2500-kva.,  transformer  specified  in  prob. 
5,  Art.  64,  had  a  total  impedance  drop  of  73.5  volts  at  full  load  current 
on  the  low-tension  side.  What  average  force  is  exerted  on  each  coil- 
face  during  a  short-circuit,  provided  that  the  line  voltage  remains 
constant?  The  transformer  winding  consists  of  12  high-tension  coils  of 
100  turns  each,  and  of  11  low-tension  coils  interposed  between  the  high- 
tension  coils,  together  with  2  half-coils  at  the  ends.  Two  of  the  dimen- 
sions of  the  coils  are  repeated  here:  Om=2.6  m.;  J=18  cm.  Hint:  The 
short-circuit  current  is  equal  to  2000/73.5  =  27.2  times  the  rated  current. 

Ans.    About  22.  metric  tons. 

Prob.  16.  What  is  the  mechanical  pressure  on  the  surface  of  the 
conductors  in  prob.  3? 

Ans.    47.6  and  5.8  milligrams  per  sq.cm. 

71.  The  Determination  of  the  Mechanical  Forces  by  Means 
of  the  Principle  of  Virtual  Displacements.  In  order  to  determine 
the  mechanical  force  or  torque  between  two  parts  of  a  mag- 
netic circuit  the  general  method  consists  in  giving  these  parts 
an  infinitesimal  relative  displacement  and  applying  the  law  of 

1  J.  K.  Sumec,  Berechnung  des  einseitigen  magnetischen  Zuges  bei  Excen- 
trizitat,  Zeitschrift  fur  Elektrotechnik  (Vienna),  Vol.  22  (1904),  p.  727.  This 
periodical  is  continued  now  under  the  name  of  Electrotechnik  und  Maschin- 
enbau. 


250  THE  MAGNETIC  CIRCUIT  [ART.  71 

the  conservation  of  energy  to  this  displacement.  From  the 
equation  so  obtained  the  component  of  the  force  in  the  direction 
of  the  displacement  can  be  calculated.  Taking  other  displace- 
ments in  different  directions,  a  sufficient  number  of  the  com- 
ponents of  the  forces  are  determined  to  enable  one  to  calculate 
the  forces  themselves.  Since  the  forces  in  a  given  position  of 
the  system  are  perfectly  definite,  the  result  is  the  same  no  matter 
what  displacements  are  assumed,  provided  that  these  displace- 
ments are  possible,  that  is,  consistent  with  the  given  conditions 
of  the  problem.  Therefore  displacements  are  selected  which 
give  the  simplest  formula?  for  the  energies  involved.  We  have 
had  two  applications  of  this  principle  in  the  preceding  article, 
in  deriving  the  expressions  for  the  tension  and  the  compression 
in  the  field,  by  giving  the  simple  magnetic  circuit  the  proper 
"  virtual  "  displacements.  In  applying  this  method,  not  only 
the  mechanical  displacement  has  to  be  specified,  but  also  the 
electric  and  the  magnetic  conditions  of  the  circuit,  in  order  to 
make  the  energy  relations  entirely  definite.  Thus,  in  the  pre- 
ceding article,  the  electromagnetic  condition  was  0  =  const.1 

First  let  us  take  the  case  when  the  partial  linkages  are  neg- 
ligible; then  according  to  the  third  eq.  (99),  the  stored  energy  is 

,    .....     .    .     (176) 


where  (ft  is  the  reluctance  of  the  circuit.  Let  F  be  the  unknown 
mechanical  force  between  two  parts  of  the  magnetic  circuit  at 
a  distance  s,  and  let  one  part  of  the  system  be  given  an  infini- 
tesimal displacement  ds.  Let  F  be  considered  positive  in  the 
direction  in  which  the  displacement  ds  is  positive.  The  mechan- 
ical work  done  is  then  equal  to  Fds.  As  in  the  preceding  article, 
let  this  displacement  take  place  with  a  constant  flux,  so  that 
there  is  no  interchange  of  energy  between  the  magnetic  circuit 
under  consideration  and  the  electric  circuit  by  which  it  is  excited. 
Then  the  work  is  done  entirely  at  the  expense  of  the  stored  energy 
of  the  magnetic  circuit,  and  we  have  : 

Fds=dWm=-dW8,      .....     (177) 
where  dWm  is  the  mechanical  work  done.     The  sign  minus  before 

1  The  principle  of  virtual  displacements  is  much  used  nowadays  in  the 
theory  of  elasticity  and  in  the  calculation  of  the  mechanical  stresses  in  the 
so-called  statically-indeterminate  engineering  structures. 


CHAP.  XIII]        TORQUE  AND  TRACTIVE  EFFORT  251 

dW8  is  necessary  because  the  stored  energy  decreases.     From 
eqs.  (176)  and  (177)  we  get 

F=  -%<P2.d(R/ds  .......     (178) 

In  some  cases  it  is  more  convenient  to  express  F  through  M 
and  (P.     We  have 


or 

F=+%M2d(P/ds.   ......     (179) 

In  the  preceding  formulae  F  is  in  joulecens,  M  is  in  ampere- 
turns,  ^  is  in  webers,  (P  is  in  henrys,  and  (ft  in  yrnehs.  With 
other  units  the  formulae  contain  an  additional  numerical  factor. 
It  is  to  be  noted  that  the  mechanical  forces  are  in  such  a  direc- 
tion that  they  tend  to  increase  the  permeance  and  decrease  the 
reluctance  of  the  circuit.  This  agrees  with  previous  statements. 
See  Arts.  41  and  57. 

If  the  partial  linkages  are  of  importance,  it  is  convenient 
to  express  the  stored  energy  in  the  form  TFs=Ji2!/,  because  the 
inductance  L  takes  account  of  the  partial  linkages;  see  eqs. 
(105)  and  (106),  Art.  58.  The  energy  equation,  according  to 
eq.  (177),  is  then 

-dW8  =  -%d(v*L)  .....     (180) 


and  the  condition  that  there  is  no  interchange  of  energy  with 
the  line  is 

d(iL)  =  Q.    .      ......     (181) 

The  latter  equation  becomes  clear  by  reference  to  eq.  (106a), 
because  Li=ntf>eq,  where  ^  is  the  equivalent  flux  under  the 
supposition  of  no  partial  linkages.  The  condition  that  there 
shall  be  no  e.m.f.  induced  in  the  winding  during  the  displace- 
ment, is  d(n<t>eg)/dt=Q,  whence  eq.  (181)  follows  directly. 

Performing  the  differentiations  in  eqs.  (180)  and  (181),  and 
substituting  the  value  of  Ldi  from  the  second  equation  into  the 
first,  we  get  that 

(182) 


When  there  are  no  partial  linkages,  L  =  n2(P,    and  eq.  (182) 
becomes  identical  with  (179). 


252  THE  MAGNETIC  CIRCUIT  [ART.  71 

Formulae  for  the  average  force  of  direct  current  electromagnets. 
In  a  tractive  magnet  (Fig.  60)  or  a  rotary  magnet  (Fig.  63) 
it  is  often  required  to  know  the  average  pull  over  a  finite  travel 
of  the  moving  part.  For  the  average  force,  the  equation 
FavedS=4Wm=  —  4W8  holds,  which  is  analogous  to  eq.  (177)  ; 
the  only  difference  being  that  finite  instead  of  infinitesimal  incre- 
ments are  used.  If  the  motion  takes  place  at  a  constant  flux, 
or  at  least  the  values  of  the  flux  are  the  same  in  the  initial  and 
the  end-positions,  we  get  from  the  preceding  formulae : 

•      V      .       (183) 

-Sl);  .     .     (184) 
-s1).    ....     (185) 

The  finite  travel  of  the  plunger  often  takes  place  at  a  con- 
stant current,  for  instance,  in  the  regulating  mechanism  of  a 


FIG.  63. — A  rotary  electromagnet. 

series  arc-lamp,  also  approximately  in  a  direct-current  electro- 
magnet connected  across  a  constant-potential  line.  Under  such 
conditions  the  foregoing  formulae  are  not  directly  appliable, 
because  they  have  been  deduced  under  the  assumption  of  no 
interchange  of  energy  between  the  electric  and  the  magnetic 
circuits,  so  that  this  case  has  to  be  considered  separately. 

Let  the  motion  be  in  the  direction  of  the  magnetic  attraction, 
and  let  the  current  remain  constant  during  the  motion.  The 
stored  energy  is  larger  in  the  end-position  than  in  the  initial 
position,  because  the  flux  is  larger  and  the  current  is  the  same. 
Therefore,  the  energy  supplied  during  the  motion  from  the  line 
must  be  sufficient  to  perform  the  mechanical  work,  and  to 


CHAP.  XIII]         TORQUE  AND  TRACTIVE  EFFORT  253 

increase  the  energy  stored  in  the  magnetic  circuit.    The  increase 
in  the  stored  energy  is 


and  the  energy  supplied  from  the  line  is  calculated  as  follows: 
The  average  voltage  induced  during  the  motion  of  the  plunger 
is  eave=i(L2—Li)/t,  where  t  is  the  duration  of  the  motion.  The 
energy  supplied  from  the  line  is  therefore 


Thus,  the  energy  supplied  from  the  line  is  twice  as  large  as  the 
work  performed,  and  we  have  the  following  important  law 
(due  to  Lord  Kelvin)  :  When  in  a  singly  excited  magnetic  circuit, 
without  saturation,  a  deformation  takes  place,  at  a  constant  current, 
the  energy  supplied  from  the  line  is  divided  into  two  equal  parts, 
one  half  increasing  the  stored  energy  of  the  circuit,  the  other  half 
being  converted  into  mechanical  work. 

According  to  this  law  we  have,  for  a  constant  current  electro- 
magnet, that  the  mechanical  work  done  is  equal  to  the  increase 
in  the  energy  stored  in  the  magnetic  field.  Hence 


Thus, 

Fave=^i2(L2-Ll)/(s2-s1)'}     ......  (186) 

or,  if  the  partial  linkages  are  negligible, 

s1)')      .    .     .     (187) 
(188) 


When  a  magnet  performs  a  rotary  motion  (Eig.  63),  the 
preceding  formulae  are  modified  by  substituting  TdO  in  place  of 
Fds,  or  Tave(62—  61)  in  place  of  Fave(s2  —  si).  Here  T  is  the  torque 
in  joules  and  (62—61)  or  dO  is  the  angular  displacement  of  the 
armature  in  radians.  Or  else,  in  the  foregoing  formulas  F  may 
be  understood  to  stand  for  the  tangential  force,  and  the  dis- 
placement to  be  ds=r  dd,  where  r  is  the  radius  upon  which 
the  force  F  is  acting.  Then  the  torque  is  T=  Fr.  If,  for  instance, 
we  apply  eq.  (179)  to  a  rotary  motion,  it  becomes 

T=Fr=+%M2d(P/dO.    '  .....     (189) 
The  other  equations  may  be  written  by  analogy  with  this  one. 


254  THE  MAGNETIC  CIRCUIT  [ART.  71 

Alternating-Current  Electromagnets.  The  preceding  formulae  are 
deduced  under  the  supposition  that  the  magnetic  field  is  excited 
by  a  direct  current.  They  are,  however,  applicable  also  to  alter- 
nating-current electromagnets,  because  the  pulsations  in  the 
current  merely  cause  the  energy  to  surge  to  and  from  the  magnetic 
circuit,  without  any  net  effect,  so  far  as  the  average  stored  energy 
and  the  mechanical  work  are  concerned.  The  average  stored 
energy  corresponds  to  the  effective  values  of  the  current  and  the 
flux. 

In  practice  two  types  of  alternating-current  electromagnets 
are  of  importance,  namely,  those  operating  at  a  constant  voltage 
and  those  operating  at  a  constant  current.  As  an  example  of 
the  first  class  may  be  mentioned  the  electromagnets  used  for 
the  operation  of  large  switches  at  a  distance  (remote  control); 
the  windings  of  such  electromagnets  are  usually  connected  directly 
across  the  line.  Constant-current  magnets  are  used  in  the 
operating  mechanism  of  alternating-current  series  arc-lamps.  In 
an  A.  C.  electromagnet  practically  all  of  the  voltage  drop  is 
reactive  and  hence  proportional  to  the  flux. 

In  a  constant-potential  electromagnet  the  effective  value  of 
the  equivalent  flux  is  the  same  for  all  positions  of  the  plunger 
(neglecting  the  ohmic  drop  in  the  winding).  Therefore,  formula 
(185)  holds  true.  Let  e  be  the  effective  value  of  the  constant 
voltage,  or  more  accurately  the  reactive  component  alone,  and 
let/  be  the  frequency  of  the  supply.  Then,  e=2xfLiii  =  2nfL2i2, 
so  that  the  formula  for  the  pull  becomes 

Fave  =  e(il-i2)/l^f(s2-s1)]  .....     (190) 

For  a  constant-current  A.C.  electromagnet  eq.  (186)  applies; 
introducing  again  the  reactive  volts  ei  =  2/r/L1t  and  e2  =  2nfL2i, 
we  get 

-si)].    ....     (191) 


In  both  cases  the  mechanical  work  performed  is  proportional 
to,  the  difference  in  the  reactive  volt-amperes  consumed  in  the 
two  extreme  positions  of  the  moving  part.1 

1  For  further  details  in  regard  to  electromagnets  consult  C.  P.  Steinmetz, 
Mechanical  Forces  in  Magnetic  Fields,  Trans.  Amer.  Inst.  Elec.  Engs.,  Vol. 
30  (1911),  and  the  discussion  following  this  paper;  also  C.  R.  Underbill 
Solenoids,  Electromagnets,  and  Electromagnetic  Windings  (1910),  chapters  6 


CHAP.  XIII]         TORQUE  AND  TRACTIVE  EFFORT  255 

Prob.  16.  Derive  formula  (171)  for  the  lifting  magnet  by  means 
of  the  principle  of  virtual  displacements.  Solution:  The  reluctance 
of  the  air-gap  is  (R  =  s/(/*A),  so  that  d(R/ds  =  l/(/*A),  hence,  according 
to  eq.  (178)  F=  -J^VCM)  =  -  AB2/2p.  The  minus  sign  indicates  that 
the  stress  is  one  of  tension. 

Prob.  17.     Derive  expression  (172)  from  formula  (179). 

Prob.  18.  Derive  the  formula  for  the  repulsion  between  the  wingings 
in  a  transformer  from  eq.  (182). 

Prob.  19.  Derive  from  eq.  (182)  the  force  of  repulsion  between  two 
infinitely  long,  parallel,  cylindrical  conductors  placed  at  a  distance 
of  b  meters  apart,  and  forming  an  electric  circuit  (Fig.  47). 

Ans.    2Mi2(l/fy  X10~8  kg.  for  I  meters  of  the  loop. 

Prob.  20.  What  deformation  of  the  windings  may  be  expected 
during  a  severe  short-circuit  of  a  core-type  or  a  cruciform  type  trans- 
former (Figs.  12  and  14)  with  cylindrical  coils,  (a)  when  the  centers 
of  the  coils  are  on  the  same  horizontal  line,  and  (6)  when  one  of  the 
windings  is  mounted  somewhat  higher  than  the  other  ? 

Prob.  21.  Show  that  in  a  constant-current  rotary  magnet  (Fig. 
63)  Ta=  Const. — that  is,  the  torque  in  the  different  positions  of  the 
armature  is  inversely  proportional  to  the  air-gap  at  the  entering  pole- 
tip.  Hint:  cKP  =  pwrdO/a,  where  w  is  the  dimension  parallel  to  the 
shaft. 

Prob.  22.  State  Kelvin's  law  when  mechanical  work  is  done  against 
the  forces  of  the  magnetic  field. 

Prob.  23.  A  60-cycle,  8-amp.,  series  arc-lamp  magnet  has  a  stroke 
of  32  mm.;  the  reactive  voltage  consumed  in  the  initial  position  is  9 
v.,  and  in  the  final  position  20  v.  What  is  the  average  pull  ? 

Ans.     372  grams. 

72.  The  Torque  in  Generators  and  Motors.  The  magnetic 
circuits  considered  in  the  preceding  articles  of  this  chapter  are 
singly  excited,  that  is,  they  have  but  one  exciting  electric  circuit. 
From  the  point  of  view  of  mechanical  forces  this  also  applies 
to  each  air-gap  in  a  transformer,  because,  neglecting  the  mag- 
netizing current,  the  primary  and  the  secondary  coils  may  be 
combined  into  equivalent  leakage  coils  (Art.  64).  On  the  other 
hand,  a  generator  or  a  motor  under  load  has  a  doubly-excited 
magnetic  circuit,  the  useful  field  being  linked  with  both  the 
field  and  the  armature  windings.  The  two  m.m.fs.  not  being 
in  direct  opposition  in  space,  the  flux  is  deflected  from  the  shortest 

to  9  incl.;  S.  P.  Thompson,  On  the  Predetermination  of  Plunger  Electro- 
magnets, Intern.  Elect.  Congress,  St.  Louis,  1904,  Vol.  1,  p.  542;  E.  Jasse, 
Ueber  Elektromagnete,  Elecktrotechnik  und  Maschinenbau,  Vol.  28  (1910), 
p.  833;  R.  Wikander,  The  Economical  Design  of  Direct-current  Electro- 
magnets; Trans.  Amer.  Inst.  Elect.  Engs.,  Vol.  30  (1911). 


256 


THE  MAGNETIC  CIRCUIT 


[ART.  72 


o 


path,  and  the  torque  is  due  to  the  tendency  of  the  tubes  of  force 
to  shorten  themselves  longitudinally,  and  to  spread  laterally. 

Consider  the  simplest  generator  or  motor,  consisting  of  a 
very  long  straight  conductor  which  can  move  at  right  angles 
to  a  uniform  magnetic  field  of  a  density  B  (Fig.  64).  Let  the 
ends  of  the  conductor  slide  upon  two  stationary  bars  through 
which  the  current  is  conducted  into  a  load  circuit  in  the  case  of 
a  generator  action,  and  through  which  the  power  is  supplied 
in  the  case  of  a  motor  action.  If  the  magnetic  circuit  contains 

no  iron,  the  resultant  field  is  a 
superposition  of  the  original  uniform 
field  and  of  the  circular  field  created 
by  the  current  in  the  conductor. 
With  the  direction  of  the  current 
indicated  in  Fig.  64,  the  resultant 
field  is  stronger  on  the  left-hand  side 
of  the  conductor  than  it  is  on  the 
right-hand  side,  and  there  is  a 
resultant  lateral  pressure  exerted 
upon  the  conductor  to  the  right. 
In  the  case  of  a  motor  action  the 
direction  of  the  motion  is  in  the 
same  direction  as  the  pressure  from 
the  stronger  field.  In  the  case  of  a 
generator  action  the  conductor  is 
moved  by  an  external  force  against  this  pressure.  Compare 
also  the  rule  given  in  Art.  24. 

To  find  the  mechanical  force  between  the  conductor  and  the 
field,  we  will  apply  again  the  principle  of  virtual  displacements. 
Let  the  current  through  the  conductor  be  i,  and  let  the  con- 
ductor be  moved  against  the  magnetic  pressure  by  a  small  amount 
ds.  Assume  that  the  stored  magnetic  energy  of  the  electric 
circuit  to  which  the  conductor  belongs  is  the  same  in  the  various 
positions  of  the  conductor.  (Such  is  the  case  in  actual  machines.) 
Then  the  work  done  by  the  external  force  is  entirely  converted 
into  electrical  energy,  and  we  have 


Gen 


Mdtor 


FIG.  64. — A  straight  conductor 
in  a  uniform  magnetic  field. 


(192) 


where  e  is  the  induced  e.m.f.     Let  both  F  and  e  refer  to  a  length 


CHAP.  XIII]         TORQUE  AND  TRACTIVE  EFFORT  257 

I  of  the  conductor.  Substituting  the  value  of  e  from  eq.  (27), 
Art.  24,  we  have 

Fds/dt=iBlv, 
or,  since  ds/dt=v, 

F=iBl.    .    .  \    .    .    .     .     (193) 

In  this  expression  i  is  in  amperes,  B  is  in  webers  per  sq.cm., 
I  is  in  cm.,  and  F  is  in  joulecens.  With  other  units  the  formula 
contains  a  numerical  factor. 

Formula  (193)  maybe  used  also  with  anon-uniform  field,  and 
also  when  the  direction  of  the  conductor  is  not  at  right  angles 
to  that  of  the  line  of  force.  In  such  cases  the  formula  becomes 
dF=iBdl,  where  dF  is  the  force  acting  upon  an  infinitesimal 
length  dl  of  the  conductor,  and  B  is  understood  to  be  the  com- 
ponent of  the  actual  flux  density  perpendicular  to  dl. 

As  an  application  of  formula  (193),  consider  the  attraction 
or  the  repulsion  between  two  straight  parallel  conductors  carrying 
currents  i\  and  1*2,  and  placed  at  a  distance  b  from  each  other. 
The  circuit  of  each  conductor  may  be  considered  closed  through 
a  concentric  cylindrical  shell  of  infinite  radius,  as  in  Art.  60. 
It  is  apparent  from  symmetry  that  the  field  produced  by  each 
system  gives  no  resultant  force  with  the  current  in  the  same 
system.  Thus,  the  mechanical  force  is  due  to  the  action  of  the 
field  1  upon  the  current  2,  and  vice  versa. 

The  flux  density  due  to  the  system  2  at  a  distance  b  from 
the  conductor  2  is  B=  /jL.i2/2xb,  so  that,  according  to  eq.  (193), 

F=fiiii2l/ 2nb  joulecens, (194) 

where  n=  1.257 X  10~8.     In  kilograms  the  same  formula  is 

F  =  2Mili2(l/b)W-8, (194a) 

provided  that  I  and  b  are  measured  in  the  same  units.  The 
force  is  an  attraction  or  a  repulsion  according  to  whether  the 
two  currents  are  flowing  in  the  same  or  in  the  opposite  directions. 
When  ii  =  i2,  this  formula  checks  with  that  given  in  prob.  19 
above. 

Formula  (193)  applies  also  to  the  tangential  force  between 
the  field  and  armature  conductor  in  any  ordinary  generator  or 
motor,  provided  that  (a)  the  conductors  are  placed  upon  a 
smooth-body  armature,  and  (6)  the  conductors  are  distributed 


258  THE   MAGNETIC  CIRCUIT  [ART.  72 

uniformly  over  the  armature  periphery,  so  that  the  stored  mag- 
netic energy  is  the  same  in  all  positions  of  the  armature.  It 
would  be  entirely  wrong,  however,  to  apply  this  formula  to 
a  slotted  armature,  using  for  B  the  actual  small  flux  density 
in  the  slot  within  which  the  conductor  lies.  This  would  give 
the  force  acting  upon  the  conductor  itself,  and  tending  to  press 
it  against  the  adjacent  conductor  or  against  the  side  of  the  slot; 
but  the  actual  tangential  force  exerted  upon  the  armature  as 
a  whole  is  many  times  greater,  and  practically  all  of  it  is  exerted 
directly  upon  the  steel  laminations  of  the  teeth. 

At  no-load,  the  flux  distribution  in  the  active  layer  of  the 
machine  is  symmetrical  with  respect  to  the  center  line  of  each 
pole  (Fig.  24),  so  that  the  resultant  pull  along  the  lines  of  force 
is  directed  radially.  The  armature  currents  distort  the  field 
as  a  whole,  and  also  distort  it  locally  around  each  tooth,  the 
general  character  of  distortion  being  shown  in  Fig.  36.  The 
unbalanced  pull  along  the  lines  of  force  has  a  tangential  com- 
ponent which  produces  the  armature  torque.  This  torque, 
although  caused  by  the  current  in  the  armature  conductors, 
is  largely  exerted  directly  upon  the  teeth,  because  the  flux  density 
there  is  much  higher. 

Thus,  in  order  to  determine  the  total  electromagnetic  torque 
in  a  slotted  armature,  it  is  again  necessary  to  apply  the  principle 
of  virtual  displacements.  The  reluctance  of  the  active  layer 
per  pole  varies  somewhat  with  the  position  of  the  armature, 
so  that  the  energy  stored  in  the  field  is  also  slightly  fluctuating. 
It  is  convenient,  therefore,  to  take  a  displacement  which  s  a 
multiple  of  the  tooth  pitch,  in  order  to  have  the  same  stored 
energy  in  the  two  extreme  positions.  This  gives  the  average 
electromagnetic  torque. 

(a)  The  Torque  in  a  Direct-current  Machine.  Let  the  virtual 
displacement  be  equal  to  6  geometric  degrees  and  be  accomplished 
in  t  seconds.  Then  we  have 

TaveO=iEt, .     (195) 

where  T  is  the  torque,  i  is  the  total  armature  current,  and  E  is 
the  total  induced  e.m.f.  Eq.  (195)  states  the  equality  of  the 
mechanical  work  done  and  of  the  corresponding  electrical  energy 
supplied.  The  average  induced  e.m.f.  is  independent  of  the 
flux  distribution,  or  of  the  presence  or  absence  of  teeth  (see 


CHAP.  XIII]        /TORQUE  AND  TRACTIVE  EFFORT  259 

Art.  24  and  prob.  18  in  Art.  26).  Take  6  to  correspond  to 
two  pole  pitches,  or  6=2n/(%p);  then  t=l/f,  where  /  is  the 
frequency  of  the  magnetic  cycles.  Substituting  these  values  and 
using  the  value  of  E  from  eq.  (37),  Art.  31,  we  get,  after 
reduction, 

Tave=iNp$/n  joules, (196) 

where  $  is  in  webers;  or 

Tave  =  0.0325iNp& X  10~2  kg-meters,     .     .     (196o) 

^  being  in  megalines. 

This  formula  does  not  contain  the  speed  of  the  machine, 
the  torque  depending  only  upon  the  armature  ampere-turns  iN 
and  the  total  flux  pti>.  Consequently,  the  formula  can  be  used 
for  calculating  the  starting  torque  or  the  starting  current  of 
a  motor.  Eqs.  (196)  and  (196a)  give  the  total  electromagnetic 
torque,  part  of  which  serves  to  Overcome  the  hysteresis,  eddy 
currents,  friction  and  windage.  The  remainder  is  available  on 
the  shaft.  When  calculating  the  starting  torque,  it  is  necessary 
to  take  into  account  the  effort  required  for  accelerating  the 
revolving  masses. 

(b)  The  Torque  in  a  Synchronous  Machine.    The  equation  of 
energy  is 

Taved=miEeos<j>'.t, (197) 

where  m  is  the  number  of  phases,  i  and  E  are  the  effective  values 
of  the  armature  current  and  the  induced  voltage  per  phase,  and 
$  is  the  internal  phase  angle  (Fig.  37).  Taking  again  a  dis- 
placement over  two  poles  and  using  the  value  of  E  from  eq.  (3) 
Art.  26,  we  get 

Tave= Q.Q36lkbmiN  cos  <£'p01(r2  kg-meters.      .     (198) 

(c)  The  Torque  in  an  Induction  Machine.     The  torque  being 
exerted  between  the  primary  and  the  secondary  members  of 
an  induction  machine,  it  may  be  considered  from  the  point  of 
view  of  either  member.     This  is  because  the  torque  of  reaction 
upon  the  stator  is  equal  and  opposite  to  the  direct  torque  upon 
the  rotor.     For  purposes  of  computation  it  is  more  convenient 
to  consider  the  torque  from  the  point  of  view  of  the  primary 
winding,  in  order  to  be  able  to  use  the  primary  frequency  and 


260  THE  MAGNETIC  CIRCUIT  [ART.  72 

the  synchronous  speed.  Therefore  eq.  (198)  gives  the  torque  of 
an  induction  machine  (including,  as  before,  friction  and  hysteresis), 
where  the  various  quantities  refer  to  the  stator.  However,  these 
quantities  may  equally  well  be  taken  in  the  secondary,  but  in 
this  case,  since  hysteresis  occurs  mainly  in  the  stator,  the  torque 
to  overcome  hysteresis  is  not  included. 

Formula  (198)  is  hardly  ever  used  in  practice,  especially 
for  the  computation  of  the  starting  torque,  because  it  is  difficult 
to  eliminate  the  large  leakage  flux  which  gives  no  torque.  It 
is  much  more  convenient  to  determine  the  torque  from  the  circle 
diagram,  or  from  the  equivalent  electric  circuit. 

In  case  the  torque  is  determined  from  the  equivalent  electric  cir- 
cuit, we  can  write  from  eqs.  (195)  and  (197)  the  expressions  for  the 
torque  directly,  by  substituting  for  6  its  value  2-Tu-  1-  (R.P.M.)/60. 
For  a  direct-current  machine, 

Tave  =  0.0325—^-—  Kg.-meters.     .    .    .     (199) 

TU'  (xi.i  .M..) 

Here  the  induced  e.m.f.  E  =  Et±iRa,  according  to  whether  the 
machine  is  a  generator  or  a  motor;   Et  is  the  terminal  voltage 
and  Ra  the  resistance  of  the  armature,  brushes,  and  series  field. 
For  an  alternating-current  machine, 

.    .    .    (200) 


In  a  synchronous  machine  E  is  the  induced  e.m.f.  and  <ft  is 
the  internal  phase  angle.  In  an  induction  machine,  iE  cos  <£' 
is  the  power  per  phase  delivered  to  the  rotor,  that  is,  the  input 
minus  hysteresis  and  primary  PR  loss.  The  term  (R.P.M.)  is 
in  all  cases  the  synchronous  speed  of  the  machine. 

Prob.  24.  Two  single-conductor  cables  from  a  direct-current  machine 
are  installed  parallel  to  each  other  at  a  distance  of  16  cm.  between  their 
centers,  on  transverse  supports  spaced  80  cm.  apart.  The  rated  current 
through  the  cables  is  850  amp.  What  is  the  force  acting  upon  each 
support  under  the  normal  conditions,  and  when  the  current  rises  to 
twenty  times  its  rated  value  during  a  short-circuit? 

Ans.    0.0737  and  29.5  kg. 

Prob.  25.  A  4-pole,  series  direct-current  motor  must  develop  a 
starting  torque  of  74  kg.-m.  (including  the  losses).  The  largest  possible 
flux  per  pole  is  about  2.5  ml.  ;  there  are  240  turns  in  series  between  the 
brushes.  Calculate  the  starting  current.  Ans.  95  amp. 


CHAP.  XIII]         TORQUE  AND  TRACTIVE  EFFORT  261 

Prob.  26.  Explain  the  reason  for  which  the  compensating  winding 
(Art.  54),  while  removing  the  armature  reaction,  does  not  affect  appre- 
ciably the  useful  torque  of  the  motor.  Hint:  this  can  be  shown  by 
applying  the  method  of  virtual  displacements;  also  from  the  fact  that 
the  local  distortion  of  the  flux  in  the  teeth  is  not  removed. 

Prob.  27.  On  the  basis  of  Art.  15,  explain  the  mechanism  by  which 
the  phenomenon  of  hysteresis  in  an  armature  core  causes  an  opposing 
torque.  Explain  the  same  for  eddy  currents. 

Prob.  28.  Demonstrate  that  formula  (193)  may  be  used  for  slotted 
armatures,  provided  that  B  stands  for  the  average  flux  density  per 
tooth  pitch.  Hint:  take  a  virtual  displacement  of  one  tooth  pitch, 
and  express  the  induced  voltage  through  the  flux  A®  =Bl\  per  tooth 
pitch. 

Prob.  29.  Show  that  in  a  single-phase  synchronous  motor  the 
torque  at  the  synchronous  speed  pulsates  at  double  the  frequency  of 
the  supply;  also  that  the  torque  is  zero  at  any  but  the  synchronous 
speed. 

Prob.  30.  Prove  that  in  an  induction  machine  the  torque  near  syn- 
chronism is  approximately  proportional  to  the  square  of  the  voltage 
and  to  the  per  cent  slip.  Hint:  i= Const.  X $  Xslip. 

Prob.  31.  Describe  in  detail  how  to  calculate  the  maximum  starting 
torque  of  a  given  induction  motor,  and  how  to  calculate  the  amount 
of  secondary  resistance  necessary  for  a  prescribed  torque. 


APPENDIX  I 
THE  AMPERE-OHM  SYSTEM  OF  UNITS 

THE  ampere  and  the  ohm  can  be  now  considered  as  two 
arbitrary  fundamental  units  established  by  an  international 
agreement.  Their  values  can  be  reproduced  to  a  fraction  of  a 
per  cent  according  to  detailed  specifications  adopted  by  practically 
all  civilized  nations.  These  two 'units,  together  with  the  centi- 
meter and  the  second,  permit  the  determination  of  the  values  of 
all  other  electric  and  magnetic  quantities.  The  units  of  mass  and 
of  temperature  do  not  enter  explicitly  into  the  formulae,  but  are 
contained  in  the  legal  definition  of  the  ampere  and  of  the  ohm. 
The  dimension  of  resistance  can  be  expressed  through  those  of 
power  and  current,  according  to  the  equation  P=I2R,  but  it 
is  more  convenient  to  consider  the  dimension  of  R  as  fundamental 
in  order  to  avoid  the  explicit  use  of  the  dimension  of  mass  [M]. 

For  the  engineer  there  is  no  more  a  need  of  using  the  electro- 
static or  the  electromagnetic  units;  for  him  there  is  but  one 
ampere-ohm  system,  which  is  neither  electrostatic  nor  electro- 
magnetic. The  ampere  has  npt  only  a  magnitude,  but  a  physical 
dimension  as  well,  a  dimension  which  with  our  present  knowledge 
is  fundamental,  that  is,  it  cannot  be  reduced  to  a  combination 
of  the  dimensions  of  length,  time,  and  mass  (or  energy).  Let 
the  dimensions  of  current  be  denoted  by  [I]  and  that  of  resistance 
by  [R] ;  let  the  dimensions  of  length  and  time  be  denoted  by  the 
commonly  recognized  symbols  [L]  and  [T]-  The  magnitudes  and 
the  dimensions  of  the  important  magnetic  units  are  expressed 
through  these  four,  as  is  shown  in  the  following  table.  For  the 
expressions  of  the  electric  and  the  electrostatic  quantities  in  the 
ampere-ohm  system  see  the  author's  "  Electric  Circuit." 

Other  units  of  more  convenient  magnitude  are  easily  created 
by  multiplying  the  above-tabulated  units  by  powers  of  10,  or 
by  adding  prefixes  milli-,  micro-,  kilo-,  mega-,  etc. 

A  study  of  the  physical  dimensions  of  the  magnetic  quantities 

262 


AMPERE-OHM  SYSTEM 


263 


is  interesting  in  itself,  and  gives  a  better  insight  into  the  nature 
of  these  quantities.  Moreover,  formulae  can  sometimes  be  checked 
by  comparing  the  physical  dimensions  on  both  sides  of  the  equa- 
tion. Let,  for  instance,  a  formula  for  energy  be  given 


where  k  is  a  numerical  coefficient.  Substituting  the  physical 
dimensions  of  all  the  quantities  on  the  right-hand  side  of  the 
equation  from  the  table  below,  the  result  will  be  found  to  have 
the  dimension  of  energy.  This  fact  adds  to  one's  assurance 
that  the  given  formula  is  theoretically  correct. 

TABLE    OF  MAGNETIC    UNITS   AND    THEIR   DIMENSIONS   IN 
THE    AMPERE-OHM    SYSTEM 


Symbol  and  Formula 

Quantity. 

Dimension. 

Name  of  the  Unit. 

M=nl 

Magnetomotive  force 

'[I] 

Ampere-turn. 

H  =  M/l 

Field     intensity,      or 

[IL-1] 

Ampere-turn    per 

m.m.f.  gradient 

centimeter. 

@=ET 

Magnetic  flux 

[IRT]* 

Weber  (maxwell). 

B  =  #/A 

Magnetic  flux  density 

[IRTL-2] 

Webers    (maxwells) 

per  square   centi- 

meter. 

(P  =  0/M 

Permeance 

[RT] 

Henry  (perm). 

(R=M/$  =  1/(P 

Reluctance 

[R-'T-1! 

Yrneh  (rel). 

Henries  (perms)  per 

v=B/H 

Permeability 

[RTL-1] 

centimeter  cube. 

v=H/B  =  l/fi 

Reluctivity 

[R-iT-iL] 

Yrnehs     (rels)     per 

centimeter  cube. 

W  =  \M® 

Magnetic    energy    or 

[I2RT] 

Joule   or   watt-sec- 

work 

ond. 

W/V=±BH 

Density  of    magnetic 

[PRTL-3] 

Joules  per  cubic 

energy 

centimeter. 

F  =  W/l 

Force 

PRTL-1] 

Joulecen. 

*  This  is  also  the  dimension  of  the  magnetic  pole  strength.  The  concept 
of  pole  strength  is  of  no  use  in  electrical  engineering,  and,  in  the  author's 
opinion,  its  usefulness  in  physics  is  more  than  doubtful.  The  whole  theory 
of  electromagnetic  phenomena  can  and  ought  to  be  built  up  on  the  two 
laws  of  circuitation,  as  has  been  done  by  Oliver  Heaviside  in  his  Electro- 
magnetic Theory. 

A  small  irregularity  is  due  to  the  use  of  the  maxwell  and  of 
its  multiples  instead  of  the  weber.  As  long  as  this  usage  persists 


264  THE  MAGNETIC  CIRCUIT 

it  is  convenient  to  use  the  corresponding  units  for  reluctance 
and  permeance,  to  which  the  author  has  ventured  to  give  the 
names  of  rel  and  perm.  Since  one  maxwell  is  equal  to  1/108  of  a 
weber,  one  perm  is  equal  to  1/108  of  one  henry,  and  one  rel  is 
108  yrnehs.  Accordingly,  permeabilities  and  reluctivities  are 
measured  in  perms  per  centimeter  cube  and  in  rels  per  centimeter 
cube  respectively. 

In  order  not  to  break  with  the  established  usage,  the  maxwell, 
the  perm,  and  their  multiples  are  employed  in  numerical  compu- 
tations in  this  book,  while  the  weber  and  the  henry  are  used  in 
the  deduction  of  the  formula,  being  the  natural  fundamental 
units  of  flux  and  permeance  in  the  ampere-ohm  system.  It  is  pos- 
sible that  the  constant  necessity  for  multiplying  or  dividing  results 
by  10~8,  due  to  the  use  of  the  maxwell,  may  prove  to  be  more  and 
more  of  an  inconvenience  in  proportion  as  magnetic  computations 
come  into  common  engineering  practice.  Then  the  weber,  the 
henry,  and  their  submultiples  will  be  found  ready  for  use,  and 
the  system  of  magnetic  units  will  be  completely  coordinated. 

Another  irregularity  in  the  system  as  outlined  above  is  caused 
by  the  use  of  the  kilogram  as  the  unit  of  force,  because  it  leads  to 
two  units  for  energy  and  torque,  viz.,  the  kilogram-meter  and 
the  joule;  1  kg.-meter=  9.806  joules.  Force  ought  to  be  measured 
in  joules  per  centimeter  length,  to  avoid  the  odd  multiplier.  Such 
a  unit  is  equal  to  about  10.2  kg.,  and  could  be  properly  called 
the  joulecen  (=107  dynes).  There  is  not  much  prospect  in  sight 
of  introducing  this  unit  of  force  into  practice,  because  the  kilo- 
gram is  too  well  established  in  common  use.  The  next  best 
thing  to  do  is  to  derive  formulae  and  perform  calculations,  whenever 
convenient,  in  joulecens,  and  to  convert  the  result  into  kilo- 
grams by  multiplying  it  by  g=  9.806.  This  is  done  in  some 
places  in  this  book. 

Thus,  leaving  aside  all  historical  precedents  and  justifications, 
the  whole  system  of  electric  and  magnetic  units  is  reduced  to 
this  simple  scheme :  In  addition  to  the  centimeter,  the  gram,  the 
second  and  the  degree  Centigrade,  two  other  fundamental  units 
are  recognized,  the  ohm  and  the  ampere.  All  other  electric  and 
magnetic  units  have  dimensions  and  values  which  are  con- 
nected with  those  of  the  fundamental  six  in  a  simple  and  almost 
self-evident  manner  (see  the  table  above). 

To  appreciate  fully  the  advantages  of  the  practical  ampere- 


AMPERE-OHM  SYSTEM 


265 


ohm  system  over  the  C.G.S.  electrostatic  and  electromagnetic 
systems,  one  has  only  to  compare  the  dimensions,  for  instance, 
of  magnetomotive  force  and  of  flux  in  these  three  systems, 
as  shown  below. 


The  Ampere- 
Ohm  System. 

C.G.S.  Electro- 
magnetic System. 

C.G.S.  Electro- 
static System. 

Dimension  of  m.m.f. 
Dimension  of  flux. 

[11 

[IRT1 

Li/feT*r-y-* 
LiAfir-y 

L!M*T-2/cl 
L*M*/c-i 

APPENDIX  II 
AMPERE-TURN  vs.  GILBERT 

THE  reader  has  probably  been  taught  before  that  the  per- 
meability of  air  is  equal  to  unity  in  the  electromagnetic  C.G.S. 
system;  silent  assumption  was  then  probably  made  that  /*=! 
also  in  the  practical  ampere-ohm  system.  The  true  situation 
is,  however,  as  follows:  In  any  system  of  units  whatsoever,  the 
fundamental  equation  $  =  (fiA/l) .  M  holds  true,  being  a  mathe- 
matical expression  of  an  observed  fact.  Now  let  the  quantities 
be  expressed  in  the  ampere-ohm  system,  and  assume  the  centi- 
meter to  be  the  unit  of  length.  The  flux  is  then  expressed 
either  in  maxwells  or  in  webers,  both  of  which  are  connected 
with  the  ampere-ohm  system  through  the  volt.  The  natural 
(though  not  the  only  possible)  unit  for  the  magnetomotive  force 
is  one  ampere-turn.  Therefore,  all  the  quantities  in  the  fore- 
going equation  are  determinate,  and  the  value  of  /*  cannot  be 
prescribed  or  assumed,  but  must  be  determined  from  an  actual 
experiment,  the  same  as  the  electric  conductivity  of  a  metal, 
or  the  permittivity  of  a  dielectric  have  to  be  determined. 
Experiment  shows  that  /*=  1.257  when  the  maxwell  is  used  as  the 
unit  of  flux,  and  hence  ju=  1.257X  10~8  if  the  flux  is  measured  in 
webers. 

It  is  possible  to  assume  /£=  1,  provided  that  the  unit  of  mag- 
netomotive force  is  not  prescribed  in  advance.  In  this  case, 
the  unit  of  magnetomotive  force,  as  determined  from  experiment, 
comes  out  equal  to  1/1.257  of  an  ampere-turn.  This  unit  is 
called  the  gilbert,  and  it  must  be  understood  that  the  permeability 
of  non-magnetic  materials  is  equal  to  unity  only  if  the  magneto- 
motive force  is  measured  in  gilberts.  To  the  author  the  advan- 
tages of  such  a  system  for  practical  use  are  more  than  doubtful. 
In  the  first  place,  the  gilbert  is  a  superfluous  unit,  because  the 
results  of  calculations  must  after  all  for  practical  purposes  be 
converted  into  ampere-turns  in  order  to  specify  the  number  of 

266 


AMPERE-TURN  vs.  GILBERT  267 

turns  and  the  exciting  current  of  windings.  Thus,  one  would 
have  to  deal  with  two  units  of  magnetomotive  force,  the  gilbert 
and  the  ampere-turn,  one  being  about  0.8  of  the  other.  In  the 
second  place,  with  the  assumption  /*=  1  for  non-magnetic  materials 
B  becomes  numerically  equal  to  H,  which  is  a  grave  inconvenience, 
because  B  and  H  are  different  physical  quantities.  B  and  H 
have  different  physical  dimensions,  because  JJL  has  a  definite 
physical  dimension,  even  though  the  numerical  value  of  it  is 
assumed  to  be  equal  to  unity  for  air.  Therefore,  to  be  sure 
that  proper  physical  dimensions  are  preserved,  one  has  to  remem- 
ber where  /*  is  omitted  in  formula,  and  for  a  physical  interpretation 
of  results  it  is  much  more  convenient  to  have  it  there,  explicitly. 

Still  another  objection  to  using  the  gilbert  and  to  putting 
jj.  equal  to  unity  for  air  is  that  the  ratio  of  the  ampere-turn 
to  the  gilbert  is  equal  to  a  quasi-scientific  constant  47T/10. 
To  the  author's  knowledge,  there  is  no  simple,  elementary  way 
of  deducing  the  value  of  this  constant,  without  going  over 
the  whole  mathematical  theory  of  electricity  and  magnetism. 
Thus,  a  constant  is  retained  in  practical  formulae,  the  significance 
of  which  remains  a  puzzle  to  the  engineer  all  his  life.  It  is  true 
that  the  value  of  /*=  1.257  is  equal  to  the  same  4^/10  after  all; 
but  in  this  case  there  is  nothing  "absolute,"  mysterious,  or 
sacred  about  the  value  of  4^/10.  The  student  is  simply  told 
that  1.257  happens  to  be  eaual  to  4^/10  because  the  value  of 
the  ampere  was  unfortunately  so  selected.  It  is  not  necessary 
to  go  into  further  details,  because  the  historical  reasons  which 
led  to  the  selection  of  the  values  of  unit  pole  and  unit  current 
hardly  hold  at  present.  All  calculations  would  be  just  as  con- 
venient if  p  were  equal  to  2.257,  or  any  other  value,  instead  of 
1.257. 

For  these  reasons  the  author  unhesitatingly  discards  the 
gilbert  in  teaching  as  well  as  in  practice  and  uses  the  ampere- 
turn  as  the  natural  unit  of  magnetomotive  force.  The  value 
of  permeability  becomes  then  an  experimental  quantity  which 
depends  upon  the  units  selected  for  flux  and  length. 

A, 


INDEX 


Active  layer  characteristic,  definition  of 168 

Aging  of  laminations 49 

Air-gap  ampere-turns 89 

factor,  definition  of 90 

factor  for  induction  machines 98 

flux,  nature  of  its  distribution 88 

permeance  and  m.m.f.,  accurate  method 92-97 

.  simplified,  permeance  of 89 

Alternating  current  electromagnets,  tractive  effort  of 254 

machines,  torques  of 259-260 

machines.     See  also  Synchronous  and  Induction. 
Alternator.    See  Synchronous  machine. 

calculation  of  regulation 155 

definition  of  load  characteristic 141 

definition  of  regulation 156 

wave-form 72 

Ampere-Ohm  system 262 

Ampere-turns,  the  cause  of  magnetism.     See  also  M.m.f 4 

Ampere-turn, 

unit  of  m.m.f 5 

vs.  Gilbert 266 

Analogue,  mechanical,  to  hysteresis 36 

to  inductance 185 

Annealing  of  laminations 50 

Apparent  flux  density  in  teeth 101 

Armature,  eccentric,  force  upon 248 

reactance  in  synchronous  machine,  nature  and  definition  of ....   140 
See  also  Inductance. 

reaction,  definition  of 140 

in  a  direct  current  machine 163 

in  an  induction  machine 131 

in  a  rotary  converter 175 

in  a  synchronous  machine 139 

Asynchronous.     See  Induction  machines. 
Auxiliary  poles.     See  Commutating  poles. 

Axial  length  of  armature,  effective 94 

269 


270  INDEX 


Back  ampere  turns.     See  Demagnetizing,  Direct. 

Belt  leakage,  description  of 223 

Belts  of  current  in  a  D.C.  machine 163 

B-H  curves  for  iron 20-24 

relation  for  air 15 

Blondel  diagram 154-155 

Breadth  factor,  definition  of 65 

formula  for 68 

Brush  shift  in  D.C.  machines 165 

Bus-bars,  repulsion  between 248 

Cable,  concentric,  flux  distribution  in 189 

induction  of  a  single  phase 191 

Carter's  curve  for  fringe  permeance 95 

Castings 21 

Characteristic,  active  layer,  definition  of 168 

air 231 

load,  definition  of  in  alternator 141 

open  circuit.     See  Saturation  curves,  and  Exciting  cur- 
rent. 

phase,  synchronous  motor 142 

Circuit,  a  simple  magnetic.     (See  also  Magnetic  circuit) .>       1 

magnetic,  containing  iron,  series-parallel 29 

definition  of 3 

Coercive  force 32 

Commutating  poles,  calculations  for.     (See  also  Interpoles) 173 

definition  of  and  location 172 

Commutation,  criterion  of 235 

description  of 233 

frequency  of 236 

of  a  fractional  pitch  winding 238 

of  a  lap  winding 235 

of  a  multiplex  winding 238 

of  a  two  circuit  winding 237 

vs.  inductance 232 

Compensating  winding  for  D.C.  machine 174 

Complete  linkages,  definition  of 181 

Compounding  in  a  D.C.  machine 170 

Compression,  lateral,  in  a  magnetic  field 244 

Computations  for  interpoles 173 

Condenser,  synchronous 157 

Conductor,  force  on,  in  a  magnetic  field • 256 

Cores  of  iron  wire 41 

of  revolving  machinery,  m.m.f .  for 105 

of  transformers,  m.m.f  for 81 

Core  loss  current,  in  an  induction  motor.     See  Core  loss  in  revolving 
machinery. 


INDEX  271 

PAGE 

Core  loss  current,  in  a  transformer 82 

curves 45 

extrapolation  of 54 

in  transformers 82 

in  revolving  machinery 46 

measurement  of 44 

or  iron  loss,  definition  of 42 

Cross  ampere-turns.     See  Reaction. 

Current.     See  also  Exciting  current. 

belts  in  a  D.C.  machine 103 

secondary,  in  induction  machine 132 

Demagnetization  and  distortion.     See  Reaction. 

Demagnetizing  reaction,  in  direct  current  machines 164 

in  synchronous  machines.     See  Direct  reaction. 

Density  of  flux,  definition  of 14 

in  teeth,  apparent 101 

Deri  winding 174 

Dimensions  of  units,  table  of 263 

Direct  current  electromagnet,  average  tractive  effort  of 252 

Direct  current  machine,  brush  shift 165, 167 

compensating  windings  for 174 

compounding  in 170 

current  belts  in 163 

demagnetizing  reaction  in 164 

e.m.f .  induced  in 75 

interpoles  in 172 

inductance  of  coils  in 236 

loaded,  field  m.m.f 170 

loaded,  flux  distribution 168 

loaded,  m.mf .  to  overcome  distortion 169 

torque , 258-260 

transverse  reaction  in 165 

Direct  current  machines.     See  also  Saturation  curves. 

windings,  vs.  induced  e.m.f 76 

vs.  inductance 237 

Direct  current  motor,  speed  under  load 171 

Direct  reaction.     See  also  Demagnetizing  reaction. 

calculation  of  coefficient  of 158 

definition  of  coefficient  of 153 

nature  of '. 150 

Displacements,  virtual,  principle  of 249 

Distortion  and  demagnetization.   See  Reaction. 

of  field  in  a  direct  current  machine,  m.m.f. -needed  to  over- 
come    169 

Distributed  windings,  advantages  and  disadvantages  of 66 

definition  of .  .  .121 


272  INDEX 

PAGE 

Distributed  windings,  for  alternator  fields 121 

e.m.f.  of.     See  E.m.f. 

,  m.m.f.  of  D.C 165 

m.m.f .  of  single-phase 123 

m.m.f.  of  polyphase 128 

Eccentric  armature,  force  upon 248 

Eddy  currents,  nature  of 40 

prevention  of 41 

separation  of,  from  hysteresis 53 

Electric  circuit,  equivalent,  torque  from 260 

Electric  loading,  specific,  definition  of 163 

Electromagnet,  A.C.,  average  tractive  effort  of 254 

average  tractive  effort  of  D.C 252 

lifting 243 

rotary,  torque  of 253 

tractive,  force  of 248 

Electromagnetic  inertia 180 

E.m.f.,  average  reactance,   as   criterion  for   commutation 235 

formula  for  induced 57 

induced  in  A.C.  machines 65 

induced  in  D.C.  machines 75 

induced  in  an  unsymmetrically  spaced  transmission  line 205 

methods  of  inducing 55 

ratio.     (See  Ratio  of  transformation.) 

ratio  in  a  rotary  converter 78 

regulation  in  an  alternator 156 

regulation  in  a  rotary  converter 78 

vs.  inductance 185 

Energy,  density  of,  in  magnetic  field 240 

lost  in  hysteresis  cycle 38 

of  magnetic  field,  none  consumed  in 177 

stored  in  magnetic  field,  description  of 177 

stored  in  magnetic  field,  formulae  for 181 

Equivalent  electric  circuit,  torque  from 260 

permeance,  definition  of 184 

secondary  winding,  reduced  to  primary 133 

Exciting  current  in  an  induction  motor .  .  .> 130 

in  a  transformer  vs.  magnetizing  current 80 

of  a  transformer,  exciting  volt-amperes 82 

of  a  transformer,  saturated  core 84 

unsaturated  core 81 

of  machinery.     See  Saturation  curves. 

Factor,  air-gap,  definition  of 90 

in  induction  motor 98 

amplitude 84 


INDEX  273 


Factor,  breadth  factor 65 

leakage,  calculation  of 108 

leakage,  definition  of 107 

slot  factor 69 

space  factor  in  iron 41 

winding  pitch  factor 71 

Faraday's  law  of  induction 57 

Fictitious  poles  as  assumed  in  synchronous  machines 151 

Field  frames,  jn.m.f .  for 106 

Field,  magnetic  (See  also  Magnetic  field) 1 

m.m.f.  at  no  load.     See  Saturation  curves. 

m.m.f .  in  loaded  D.C.  machine 170 

loaded  synchronous  machines 148,  156 

Field  pole  leakage,  calculation  of 110 

effect  of 108 

effect  of  load  upon 113 

effect  of  saturation  upon 112 

Field  poles,  m.m.f.  for 107 

Figure  of  loss , 47 

Fleming's  rule 59 

Flux.     See  also  Leakage  flux. 

air-gap,  nature  of  its  distribution 88 

density,  definition  of 14 

in  teeth,  apparent 101 

description  of .' .  5 

distribution  of,  in  a  concentric  cable 189 

in  a  loaded  D.C.  machine 169 

in  a  loaded  synchronous  machine 139 

distortion.     See  Reaction . 

fringing,  definition  of 88 

permeance  of ; 93 

gliding  or  revolving 126 

leakage,  definition  of 16 

in  induction  machine,  description  of 221 

in  revolving  machinery,  nature  of 86 

refraction 119 

units  of 6 

Force,  lines  of  magnetic 2 

mechanical,  in  magnetic   field,  average  tractive  effort  in  D.C. 

electromagnets 252 

formulae  for  the  actual  force ....  251 

lateral  compression 244 

longitudinal  tension 242 

of  A.C.  magnets 254 

of  a  lifting  magnet 243 

of  a  tractive  magnet 248 

on  an  eccentric  armature 248 


274  INDEX 

PAGE 

Force,  mechanical,  in  magnetic  field,  on  conductors  carrying  current .  .  256 

on  transformer  coils 245 

pinch  phenomenon 246 

reason  for 242 

repulsion  between  bus-bars 248 

torque  in  a  rotary  magnet 253 

Fourier,  method  for  analyzing  waves  for  their  harmonics,  used 124,  161 

Fractional  pitch  windings,  advantages  and  disadvantages  of 67 

effect  on  inductance  in  induction  machines  .  .  225 

effect  on  inductance  in  synchronous  machines  231 

vs.  commutation  in  direct  current  machines. . .  238 

Frequency  of  commutation 236 

Fringing,  definition  of 88 

Full  load  m.m.f .  in  a  D.C.  machine 170 

in  an  induction  machine 131 

in  a  synchronous  machine 148,  156 

Gauss 14 

Generator  action 56 

Gilbert 12,  266 

Gliding  and  pulsating  m.m.f 126 

Harmonics,  of  e.m.f .,  in  alternating  current  machines 72 

vs.  voltage  ratio  in  a  rotary  coverter 79 

of  rectangular  m.m.f.  wave 123 

upper,  of  m.m.f.  in  an  induction  machine 136 

Heart,  definition  of 215 

Heating  due  to  hysteresis 38 

Henry  as  unit  of  inductance 184 

permeance,  definition  of 9 

Herring's  experiment ." 56 

Hysteresis  and  saturation,  explanation  of 34 

cycle,  energy  lost  in 34 

irreversible 34 

description  of 32 

empirical  equation  for 48 

loop 33 

vs.  heating 34 

separated  from  the  eddy  current  loss 53 

measurement  of 40 

mechanical  analogue  to 36 

Induced  e.m.f.,  formulae  for 57 

in  a  D.C.  machine 75 

in  an  alternator  and  in  an  induction  machine 65 

in  a  transformer 62 

Inductance  and  e.m.f.,  relation  between 185 


INDEX  275 


Inductance  as  electromagnetic  inertia 180 

definition  of,  and  formula}  for 184 

henry  and  perm  as  units  of 184 

mechanical  analogue  of 185 

of  a  concentric  cable,  single  phase 191 

of  circuits  in  the  presence  of  iron 186 

of  coils  and  loops 188 

of  coils  in  a  D.C.  machine 236 

of  synchronous  machines,  measurement  of,  in  A.C.  machines 

219,  231 

of  transformers,  constants  for 215 

formulae  for 211-214 

vs.  the  end  coils 213 

vs.  the  number  of  coils 213 

vs.  the  shape  of  the  coils 212 

of  transmission  lines,  single  phase 199 

three- wire  symmetrical  spacing 201 

of  windings,  formula  for 219 

fractional  pitch 225 

how  measured 219 

vs.  commutation 232 

vs.  leakage  permeance 184,  219 

Induction,  law  of 6,  57 

machines,  armature  reaction 131 

e.m.f .  induced  in 65 

higher  harmonics  in  the  m.m.f.  of  a 136 

inductance  vs.  winding  pitch -. 225 

leakage  flux,  description  of  the 221 

leakage  permeance,  values  of  the '.  225 

motor,  ratio  of  transformation  in  a 134 

secondary  current  in  a 132 

torque  in 259-260 

of  e.m.f.,  methods  of 55 

Inertia,  electromagnetic 180 

Insulator,  magnetic 17 

Intensity,  magnetic,  definition  of 13 

Interference.     See  Armature  reaction,  and  Reaction. 

Interpoles,  definition  of  (See  also  Computating  poles) 172 

Irregular  paths,  permeance  of;  how  to  map  field  in 116 

Iron,  grades  of  and  their  use 22 

laminations,  preparation  of 50 

reason  for 41 

loss,  definition  of 42 

effects  of '. 43 

figure  of  loss ". 47 

magnetization  curves 20-24 

properties  of,  general 20,  32 


276  INDEX 

PAGE 

Iron,  saturation  curves 20-24 

silicon  steel 49 

space  factor  in  laminations 41 

used  in  permanent  magnets 50 

virgin  state  of 32 

vs.  inductance 186 

wire  cores 41 

Joints  in  transformers 81 

Joulecen,  definition  of 264 

Kelvin's  law 253 

Knee  of  saturation  curve 25 

Laminations,  grades  of 44 

preparation  of 50 

reason  for 41 

Lateral  compression  in  the  magnetic  field 244 

Law  of  induction 6,  57 

Law,  Ohm's,  for  magnetic  circuit 7 

Laws  of  circulation 7,  57 

Leakage  coil,  definition  of 215 

factor,  calculation  of 108 

definition  of 107 

vs.  leakage  flux  and  permeance 109 

Leakage  flux 16 

about  armature  windings 218 

belt,  description  of 223 

in  field  poles,  as  affected  by  load 113 

as  affected  by  saturation 112 

effect  of 108 

in  induction  motors,  description  of 221 

in  transformers 208 

nature  of,  in  machinery 86 

zig-zag,  description  of 223 

Leakage  inductance.     See  Inductance. 

See  also  Leakage  permeance. 

Leakage  permeance  between  field  poles 108 

in  induction  machines,  values  of 225 

in  synchronous  machines,  values  of 230 

of  coils  in  a  D.C.  machine 236 

of  slots,  calculation  of 226 

of  windings,  formula  for . 220 

of  zig-zag  or  tooth  tip  leakage,  calculation  of 227 

Lehmann,  Dr.  Th.,  method  of  finding  permeance  of  irregular  field 18 

Lifting  magnet 243 


INDEX  277 


Lines  of  force 2 

Linkages,  fact  of 3 

as  a  measure  of  energy 180 

partial  and  complete,  definition  of 181 

Load  characteristics  of  alternator,  definition  of 141 

Loaded  D.C.  machine,  flux  distribution  in 169 

field  m.m.f.  in 170 

D.C.  motor,  speed  of  a 171 

induction  machine,  m.m.f.  relations  in 132 

synchronous  machine,  flux  distribution  in ; 139 

Loading,  specific  electric,  definition  of 163 

Longitudinal  tension  in  magnetic  field 242 

Loop,  hysteresis 33 

Machinery,  core  loss  in  revolving 46 

leakage  flux  in,  nature  of 86 

Machines,  revolving,  how  torque  is  produced  in 255,  258 

torque  in 259,  260 

types  of  synchronous 63 

Magnets,  A.C.,  average  tractive  effort  in 254 

average  tractive  effort  in  D.C 252 

lifting 243 

molecular 34 

permanent,  iron  used  in 50 

torque  of  rotary  magnets 253 

tractive,  force  of  a 248 

Magnetic  circuit,  definition  of 3 

Ohm's  law  for  the 7 

types  of,  in  revolving  machinery 85 

simple 1 

with  iron,  in  series  and  parallel 29 

Magnetic  field,  description  of 1 

density  of  energy  in 240 

energy  stored  in,  description  of 177 

formulae  for 181 

formulae  for  the  actual  force  in  a 251 

formulas  for  average  force  in  a 252 

lateral  compression  in 244 

longitudinal  tension  in 243 

of  transmission  line,  description  of 193 

shape  of 196 

reason  for  mechanical  forces  in 242 

Magnetic  insulation 17 

intensity,  definition  of 13 

relation  to  m.m.f .- 13 

potential 16 

state. .  1 


278  INDEX 

PAGE 

Magnetization  curves.     (See  also  Saturation  curves.) 20-24 

Magnetizing  current  vs.  exciting  current 80 

Magnetism,  cause  of 1,4 

M.m.f .  or  magneto  motive  force,  definition  of 5 

for  air-gap 89 

for  armature  cores 105 

for  a  D.C.  machine  to  compensate  for  distortion 168 

for  field  frames 106 

for  field  poles 107 

for  teeth,  saturated 101 

for  teeth,  tapered 100 

gliding 126 

in  induction  machines,  higher  harmonics  of 136 

of  field,  in  a  loaded  D.  C.  machine 170 

of  windings,  concentrated 121 

distributed  single  phase 123 

distributed  polyphase 128 

relations  in  an  induction  machine 132 

a  synchronous  machine 143 

wave,  harmonics  of 123,  136 

Maxwell,  definition  of 6 

Measurement  of  core  loss 44 

of  hysteresis : 40 

of  inductance 219 

Mechanical  analogue  of  hysteresis 36 

of  inductance 185 

force.     See  Force,  mechanical. 

Methods  of  inducing  e.m.f 55 

Molecular  magnets 34 

Multiplex  windings,  commutation  of 238 

Mutual  induction,  in  transmission  lines 205 

in  the  windings  of  machines 219 

See  also  Transformer  action 55 

Neutral  zone,  in  D.C.  machine 163 

Ohm's  law  for  the  magnetic  circuit 7 

Open  circuit  characteristic.     See  Saturation  curves. 

current.     See  Exciting  current. 

Overload  capacity  of  a  synchronous  motor 148 

Parallel  combination  of  permeances 16 

-series  circuits  containing  iron 29 

Partial  linkages,  definition  of 181 

Perm,  definition  of 9 

Permeability  curves 24 

definition  of .  .                                             11 


INDEX  279 

PAGE 

Permeability,  equation 24 

of  air,  discussed 266 

of  non-magnetic  materials 11 

relative 24 

vs.  reluctivity 11 

vs.  saturation 24 

Permeance,  combinations  of,  in  parallel  and  series 16 

definition  of 9 

equivalent,  definition  of 184 

leakage,  in  D.C.  machines 236 

in  induction  machines,  values  of 225 

in  synchronous  machines,  values  of 230 

measurement  of 219,  231 

of  slot,  formula  for 226 

of  windings,  formula  for 220 

zigzag,  formula  for 228 

of  air-gap,  accurate  method 92-97 

simplified 89 

of  irregular  paths 116 

of  pole  fringe 94 

of  tooth  fringe 93 

units  of *  9 

vs.  dimensions 11 

Phase  characteristics  of  a  synchronous  motor 142 

relation  of  m.m.fs.  in  an  induction  machine 132 

in  a  synchronous  machine 145 

Pinch  phenomenon 246 

Pole  face  windings,  Ryan 174 

fringe  permeance 94 

Poles,  fictitious,  assumed  in  synchronous  machines 151 

non-salient  in  a  synchronous  machine 121, 143 

salient,  definition  of 142 

Potential,  definition  of 16 

Potier  diagram 146 

Pulsating  and  gliding  m.m.fs 126 

Ratio  of  transformation 134 

of  voltages  in  a  rotary  converter 78 

Rayleigh,  Lord,  method  for  finding  permeance  of  a  field 116 

Reactance,  equivalent,  of  an  unsymmetrical  transmission  line 206 

in  synchronous  machines,  nature  of 140 

voltage,  average,  as  criterion  of  commutation 235 

Reaction.      See  also  Armature  reaction. 

demagnetizing,  in  a  D.C.  machine 164 

direct  and  transverse,  nature  of 150 

calculation  of  the  coefficient  of 158 

definition  of  the  coefficient  of .  153 


280  INDEX 

PAGE 

Reaction,  distorting.     See  Transverse  reaction. 

transverse,  calculation  of  the  coefficient  of 160 

definition  of  the  coefficient  of 153 

in  a  direct  current  machine 165 

in  a  synchronous  machine 150 

Rectangular  m.m.f .  wave,  harmonics  of 123 

Regulation  of  alternators,  calculation  of 146-8,  155-6 

definition  of 156 

Rel,  definition  of 8 

Relative  permeability 24 

Reluctance,  definition  of 7 

in  series  and  in  parallel 16 

of  irregular  paths 116 

of  various  magnetic  circuits.     See  Permeance. 

unit  of 8 

vs.  dimensions 11 

Reluctivity,  definition  of 11 

of  air 11 

vs.  permeability 11 

Remanent  magnetism.     See  Residual. 

Repulsion  between  bus-bars 248 

between  transformer  coils 245 

Residual  magnetism 32 

Resistance,  equivalent,  of  unsymmetrical  transmission  line 207 

Revolving  field 126 

machinery,  core  loss  in 46 

types  of  magnetic  circuit  in 85 

See  also  particular  kind  of  machinery. 

Right-hand  screw  rule 2 

Rotary  converter,  armature  reaction  in  a 175 

voltage  ratio  in  a 78 

voltage  regulation  in  a 78 

Ryan,  pole  face  winding  for  D.C.  machines 174 

Salient  poles,  definition  of 142 

in  synchronous  machines,  Blondel  diagram  for 154 

Saturation  and  hysteresis,  an  explanation  of 34 

and  permeability 24 

curve,  knee  of 25 

curves  of  iron 20-24 

of    machinery.       See     also     Transformers,     exciting 

current : 85-87 

effect  of,  upon  pole  leakage 112 

per  cent  of 26 

Secondary  current,  calculation  of,  in  an  induction  motor 132 

winding,  equivalent,  reduced  to  the  primary 133 

Semi-net  length 220 


INDEX  281 

PAGE 

Semi-symmetrical  spacing 203 

Series-parallel  circuits  with  iron 29 

Series-reluctances 16 

Sheet  metal.     See  Laminations. 

Short  chord  or  short  pitch  windings.     See  Fractional  pitch  windings. 

Silicon  steel 49 

Simple  magnetic  circuit 1 

Skin  effect 188 

Slot  factor,  definition  of 68 

formulae  for " 69 

leakage  permeance,  calculation  of 226 

Slotted  armature,  how  torque  is  produced  in 258 

Solenoidal,  term  applied  to  magnetic  field 6 

Space  factor  in  iron 41 

Sparking.     See  Commutation. 

Speed  of  a  variable-speed  D.C.  motor  under  load 171 

Squirel-cage  winding 133 

Steinmetz's  law  for  hysteresis 48 

Steel.     See  Iron. 

Stray  flux  or  leakage 16 

Superposition,  principle  of 194 

Synchronous  condensers 157 

Synchronous  machines,  armature  reaction  in 140 

e.m.f .  induced  in 65 

fictitious  poles  assumed  in 151 

loaded,  flux  distribution  in 139 

measurement  of  leakage  inductance 219,  231 

overload  capacity  of  motor 148 

phase  relation  of  m.m.fs.  in  a 143-145 

reactive  drop  in  a  (limits  of) 229 

torque  in 259,  260 

types  of 63 

values  of  leakage  permeance  of  the  windings  of.  .  230 
vector  diagrams  of,  with  non-salient  poles .  .    145,  147 

with  salient  poles 154 

Synchronous  motor,  definition  of  phase  characteristices  of 142 

overload  capacity  of 148 

Teeth,  apparent  flux  density  in 101 

permeance  of  tooth  fringe 93 

saturated,  m.m.f .  for 101 

tapered,  m.m.f.  for 100 

Tension.     See  also  Force,  mechanical 

longitudinal,  in  magnetic  field 242 

Torque.     See  also  Force,  mechanical. 

from  equivalent  electric  circuit. 260 

in  revolving  machines,  how  produced 258 


282  INDEX 

PAGE 

Torque  in  slotted  armatures,  how  produced . . 258 

of  rotary  magnets . 253 

of  revolving  machinery 258  260 

Tractive  effort,  actual 251 

average,  in  a  D.C.  magnet 252 

in  an  A.C.  magnet,  average 254 

magnet,  force  of 248 

Transformer  action 56 

Transformers,  core  loss  current  in  a 82 

exciting  current  in  a .' 80,  83 

induced  e.m.f .  in  a 62 

inductance  of 211-214 

as  affected  by  end  coils 213 

as  affected  by  number  of  coils 213 

as  affected  by  the  shape  of  the  coils 212 

values  of  the  constants 215 

joints  in  the  magnetic  circuit  of  a 81 

leakage  flux  in  a 208 

magnetizing  current  in  a 80 

repulsion  between  the  coils  of  a 245 

types  of 60 

Transmission  line,  description  of  the  field  of  a 193 

shape  of  the  field  calculated 196 

single  phase,  inductance  of 199 

three-wire,  symmetrical  spacing,  inductance  of 201 

unsymmetrical  spacing,  e.m.f.  induced  in 205 

unsymmetrical  spacing,  equivalent  reactance  of 206 

unsymmetrical  spacing,  equivalent  resistance  of 207 

Transverse  and  direct  reaction,  nature  of 150 

reaction,  calculation  of  the  coefficient  of 160 

definition  of  the  coefficient  of 153 

in  a  D.C.  machine 165 

in  a  synchronous  machine 150 

Turning  moment  in  machinery 258-260 

Unit  of  flux 6 

of  inductance .N 184 

of  permeance 9 

of  reluctance 8 

Units,  table  of  the  dimensions  of 263 

V  curve  or  phase  characteristics '. 142 

Variable-speed  D.C.  motor  under  load 171 

Vector  relations  in  a  synchronous  machine 145,  147,  154,  155 

with  non-salient  poles 145 

with  salient  poles 154 

Virgin  state  of  iron 32 

r 


INDEX  283 


Virtual  displacements,  principle  of 249 

Voltage.     See  E.m.f. 

Wave  form  of  e.m.f.,  effect  of  type  of  windings  upon 73 

how  to  get  sine  wave 72 

Wave-wound  armatures,  commutation  of 237 

Weber,  definition  of 6 

Wire,  iron,  for  cores 41 

Winding-pitch  factor,  values  of 71 

Windings,  compensating,  for  D.C.  machines, 174 

concentrated,  m.m.f .  of 121 

for  D.C.  machines,  effect  of  type  on  voltage 76 

distributed,  definition  of 121 

in  A.C.  machines,  advantages  and  disadvantages 

of 66 

formula  for  the  inductance  of 219 

fractional  pitch,  advantages  and  disadvantages  of 67 

commutation  of 238 

effect  of,  on  induced  e.m.f 67,  76 

on  inductance 225,  231 

inductance  and  reactance,  how  measured 219 

leakage  permeance  of,  formula  for 220 

multiplex,  commutation  of 238 

pole-face  for  D.C.  machines 174 

polyphase,  m.m.f.  of 128 

wave  or  two-circuit,  commutation  of 237 

Yrneh,  definition  of 10 

Zig-zag  leakage,  description  of 223 

permeance,  calculation  of 227 


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EN 


DEC  1 3  1947 
APR     6  1948 


MAR'  11 M949  0q •*- 
NOV  12  ][ 

rx  i  o 

MAR  1 1 1952 
APR  16  1952' 

MAR 


8    1959 


INHERING  LII3RARY 


t 


LD  21-100m-12,'46(A2012sl6)4120 


ft: 


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..ngineering 
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254299 


